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4.4. Suppose Ct is a general pencil of smooth genus g plane curves acquiring an ordinary cusp (a singularity whose local equation is given by y2 = x3). Describe the stable limit of this family of curves. Exercise 4.5. Read and do the exercises in Chapter 3 Section C of [HM]. 5. Deligne-Mumford Stacks In this section for completeness I will give you the deﬁnition of Deligne-Mumford stacks. I will summarize a few basic results and deﬁnitions. Much better accounts exist in [DM], [Ed] and [LM-B]. See also [Fan]. Let S be the category of schemes over a scheme S. A category T over S is a category together with a functor p : T � S. Deﬁnition 5.1 (Groupoid). A category ((T, p) over S is a groupoid if the following two conditions hold (1) If f : B≤ � B is a morphism in S and C is an object in T lying over B, then there exists an object C ≤ over B≤ and a morphism ζ : C ≤ � C such that p(ζ) = f . (2) Let C, C ≤, C ≤≤ be objects in T lying over the objects B, B≤, B≤≤ in S, re spectively. If ζ : C ≤ � C and ω : C ≤≤ � C are morphisms in T and f : B≤ � B≤≤ is a morphism in S satisfying p(ω) ∩ f = p(ζ), then there is a unique morphism π : C ≤ � C ≤≤ such that ω ∩ π =	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
ζ), then there is a unique morphism π : C ≤ � C ≤≤ such that ω ∩ π = ζ and p(π ) = f . Example 5.2. Recall that a Deligne-Mumford stable curve (or simply a stable curve) of genus g → 2 over a scheme S is a proper, ﬂat family β : C � S whose geometric ﬁbers are reduced, connected, one dimensional schemes Cs satisfying the following properties: (1) The only singularities of Cs are ordinary double points. (2) A non-singular rational component of Cs meets the other components in at least three points. (3) Cs has arithmetic genus g—equivalently h1(OCs ) = g. We can deﬁne a groupoid Mg of Deligne-Mumford stable curves of genus g over schemes over Spec Z as follows: The sections of Mg over a scheme X are families 13 of stable curves C � X. A morphism between C ≤ � X ≤ and C � X is a ﬁber diagram C ≤ C X which induces an isomorphism C ≤ �= X ≤ ×X C. � � X ≤ Mg is a groupoid and it is the main example that we are interested in. For the sake of future constructions and deﬁnitions it is important to keep in mind the examples of two more groupoids. Example 5.3. Any contravariant functor F : S � {sets} from schemes to sets gives rise to a groupoid (usually also called F by abuse of notation). The objects of the groupoid F are pairs (X, �) where X is a scheme and � is an element of the set F (X). A morphism between (X, �) and (Y, λ) is a	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
and � is an element of the set F (X). A morphism between (X, �) and (Y, λ) is a morphism f : X � Y such that F (f )(λ) = �. In particular, this construction allows us to view schemes as groupoids. To a scheme X we can associate its functor of points Hom(�, X). Since this is a contravariant functor from schemes to sets, to a scheme X we can also associate a groupoid X. The distinction between a scheme X and the associated groupoid is often blurred. Example 5.4. Since the construction of many moduli spaces involves taking the quotient of a parameter space (such as a component of a Hilbert scheme) by a group action, the groupoid [X/G] is important. Let X be a scheme and G a group scheme acting on X. The sections of [X/G] over a scheme Y are principal G-bundles E � Y together with a G-equivariant map E � X. A morphism between two such principal G-bundles is a pull-back diagram. Exercise 5.5. There is a relation between the previous two examples. Show that if the action of G on X is free and a quotient scheme X/G exists, then then there is an equivalence of categories between [X/G] and the groupoid associated to the scheme X/G. Let (T, p) be a groupoid. For any two objects X and Y in the ﬁber of T over a scheme B, we can associate a functor IsomB (X, Y ). This functor associates to any morphism f : B≤ � B, the set of isomorphisms in T (B≤) between f �(X)	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
� � B, the set of isomorphisms in T (B≤) between f �(X) and f �(Y ). In the case of Deligne-Mumford stable curves, given any two stable curves C and C ≤ , IsomX (C, C ≤) associates to any morphism f : Y � X the set of isomorphisms between f �(C) and f �(C ≤). Recall that C and C ≤ are both canonically polarized by �C/X and �C /X , respectively. Moreover, the formation of the relative dual izing sheaf commutes with base change. Consequently, any isomorphism satisﬁes f �(�C /X ) = �C/X . Hence, all isomorphisms are isomorphisms between polarized schemes. It follows by the existence of the Hilbert scheme, that IsomX (C, C ≤) is represented by a scheme quasi-projective over X. Deﬁnition 5.6 (Stack). A groupoid (T, p) over S is a stack if (1) IsomB (X, Y ) is a sheaf in the ´etale topology for all B, X and Y ; 14 � � (2) If {Bi � B} is a covering of B in the ´etale topology, and Xi are a collection of objects in T (Bi) with isomorphisms ζi,j : Xj\|Bi ×B Bj � Xi\|Bi ×B Bj in T (Bi ×B Bj ) satisfying the cocycle condition, then there exists an object X ⊗ T (B) with isomorphisms X\|Bi � Xi inducing the isomorphisms ζi,j . Example 5.7. The groupoid [X/G] deﬁned in Example 5.4 is a stack. Let e, e≤ be two objects in [X/G	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
/G] deﬁned in Example 5.4 is a stack. Let e, e≤ be two objects in [X/G](Y ) corresponding to two principal G-bundles E, E ≤ � Y with G-equivariant maps f, f ≤ to X, respectively. IsomY (e, e≤) is empty unless E = E≤ and f = f ≤ . In the latter case the isomorphisms correspond to the subgroup of G that stabilizes the map f . Since the functor that associates to a G-equivariant map its stabilizer is representable, condition (1) follows. Condition (2) also holds for principal G-bundles. Let Pg,n(m) be the Hilbert polynomial (2nm − 1)(g − 1), the Hilbert polynomial of an n-canonically embedded stable curve. Set N = n(2g−2)−g. Let H g,n the sub- scheme of the Hilbert scheme Hilb(2nm−1)(g−1)(PN ) parameterizing n-canonically embedded stable curves. Below we will show that there is an equivalence of cate gories between Mg and [H g,n/PGL(N + 1)] where the action of PGL(N + 1) on the Hilbert scheme is the one induced by its usual action on PN . In particular, it follows from the previous example that Mg is a stack. Recall in example 5.3 we associated to a scheme a groupoid. Observe that this groupoid is a stack. The second condition is satisﬁed because the functor of points of a scheme is represented by the scheme itself. In particular, we can view each scheme as a stack. In the litterature stacks that arise this way are usually referred to as schemes meaning that the stack associated to the scheme	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
each scheme as a stack. In the litterature stacks that arise this way are usually referred to as schemes meaning that the stack associated to the scheme. We will also indulge in this habit. A morphism of stacks f : T � T ≤ is representable if for any map of a scheme X � T ≤ the ﬁber product T ×T X is represented by a scheme. We can transport the notions of morphisms of schemes to representable morphisms of stacks in the following way: We say that a representable morphism f : T � T ≤ has a property P (such as quasi-compact, separated, proper, etc.) if for all maps of a scheme X � T ≤ , the corresponding morphism of schemes T ×T X � X has the property P . Deﬁnition 5.8 (Deligne-Mumford stack). A stack is called a Deligne-Mumford stack if (1) The diagonal ΩX : T � T ×S T is representable, quasi-compact and sepa rated; (2) There exists a scheme U and an ´etale, surjective morphism U � T . Morphisms as in condition (2) are called ´etale atlases. The following is a useful theorem for verifying that a stack is a Deligne-Mumford stack (see [DM] Theorem 4.21, or [Ed] Theorem 2.1). Theorem 5.9. Let T be a quasi-separated stack over a Noetherian scheme S. Suppose that (1) The diagonal is representable and unramiﬁed, (2) There exists a scheme U of ﬁnite type over S and a smooth, surjective S-morphism U � F . 15 The F is a Deligne-Mumford stack	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
surjective S-morphism U � F . 15 The F is a Deligne-Mumford stack. A consequence of this theorem is that if X/S is a Noetherian scheme of ﬁnite type and G/S is a smooth group scheme acting on X with with ﬁnite and reduced stabilizers, then [X/G] is a Deligne-Mumford stack. The conditions on the stabiliz ers (that they are ﬁnite and reduced) guarantee that IsomB (E, E) are unramiﬁed. It follows that the diagonal is unramiﬁed. The second condition in the theorem is satisﬁed by the map X � [X/G]. Given the equivalence of categories between Mg and [H g,n/PGL(N + 1)] it follows that Mg is a Deligne-Mumford stack because the action of PGL(N + 1) on H g,n has ﬁnite and reduced stabilizers. Just like in the case of schemes there are valuative criteria for separatedness and properness. We now state these and observe that Mg is a proper Deligne-Mumford stack. For the following two theorems let f : T � S be a morphism of ﬁnite type from a Deligne-Mumford stack to a noetherian scheme S Theorem 5.10 (The valuative criterion for separatedness). The morphism f is separated if and only if for any complete discrete valuation ring with algebraically closed residue ﬁeld and any commutative diagram T �� g1 ,g2 f S Spec R any isomorphism between the restrictions of g1 and g2 to the generic point of Spec R can be extended to an isomorphism of g1 and g2. Theorem 5.11 (The valuative criterion of properness	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
of g1 and g2. Theorem 5.11 (The valuative criterion of properness). If f is separated, then f is proper if and only if, for any discrete valuation ring R with ﬁeld of fractions K and any map Spec R � T which lifts over Spec K to a map to T , there is a ﬁnite extension K ≤ of K such that the lift extends to all of Spec R≤ where R≤ is the integral closure of R in K ≤ . The stable reduction theorem together with the valuative criterion of properness implies that Mg is a proper Deligne-Mumford stack. One approach for constructing the coarse moduli scheme (which we cannot com plete at present because we have not yet developed the theory of divisors on the moduli stack) is to ﬁrst construct the moduli space as an algebraic space, then ex hibit an ample divisor on the coarse moduli algebraic space. This approach has been applied successfully to represent many moduli functors. The ﬁrst step is achieved by a corollary of a general theorem of Keel and Mori [KM] (see also [Li] for a nice treatment). Theorem 5.12. Any separated Deligne-Mumford stack of ﬁnite type has a coarse moduli space in the category of algebraic spaces. Once we study the ample cone in the Picard group of the moduli stack, we will be able to deduce the existence of a coarse moduli scheme from the previous theorem. The second approach to the construction of the coarse moduli scheme is to directly take the G.I.T. quotient of the Hilbert scheme parameterizing n-canonically embedded stable curves. The advantage of	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
to directly take the G.I.T. quotient of the Hilbert scheme parameterizing n-canonically embedded stable curves. The advantage of the ﬁrst approach is that it does away 16 � � � with delicate calculations describing the stable and semi-stable loci of this action. The ﬁrst approach may also be used to construct moduli spaces in other situations. The advantage of the second approach is that it produces a projective coarse moduli scheme at once. 6. The GIT construction of the moduli space Good references for this section are [HM] Chapter 4, [Mum3], [FKM] and [Ne]. Explaining the GIT construction in detail would take us too far aﬁeld. Instead we will brieﬂy sketch the main ideas and refer you to the literature. 6.1. Basics about G.I.T.. An algebraic group G is a group together with the structure of an algebraic variety such that the multiplication and inverse maps are morphisms of varieties. An action of an algebraic group G on a variety X is a morphism f : G × X � X such that f (gg≤, x) = f (g, f (g≤, x)) and f (e, x) = x, where e is the identity of the group. The stabilizer of a point x ⊗ X is the closed subgroup of G ﬁxing x. The orbit of a point x under G is the image of f restricted to G × {x}. For our purposes we can always restrict attention to SL(n), GL(n) or PGL(n). An algebraic group which is isomorphic to a closed subgroup of GL(n) is called a linear algebraic group. A group is called geometrically reductive if for every linear action of G on kn and every non-zero invariant point v ⊗ kn, there exists	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
geometrically reductive if for every linear action of G on kn and every non-zero invariant point v ⊗ kn, there exists an invariant homogeneous polynomial that does not vanish on v. The group is called linearly reductive if the homogeneous polynomial may be taken to have degree one. Finally a group is called reductive if the maximal connected normal solvable subgroup is isomorphic to a direct product of copies of k� . In characteristic zero these concepts coincide. In characteristic p > 0 a threorem of Haboush guarantees that every reductive group is geometrically reductive. The question is to obtain a quotient of a variety under the action of a reductive group. Lemma 6.1. Let G be a geometrically reductive group acting on an aﬃne variety X. Let W1 and W2 be two disjoint invariant closed orbits. Then there exists an invariant polynomial f ⊗ A(X)G such that f (W1) = 0 and f (W2) = 1. Proof. Pick any h ⊗ A(X) such that h(W1) = 0 and h(W2) = 1. Consider the subspace spanned by hg for g ⊗ G. This is a ﬁnite dimensional subspace. To see this consider the function H(g, x) = h(gx) in A(G × X) � A(G) ∗ A(X). We = can write H(g, x) as a ﬁnite sum Fi ∗ Hi in A(G) ∗ A(X) of the generators of A(G) and A(X). Hence the subspace spanned by hg for g ⊗ G is contained in the subspace spanned by the Hi. Pick a basis for this subspace h1, . . . , hn. We obtain	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
in the subspace spanned by the Hi. Pick a basis for this subspace h1, . . . , hn. We obtain a rational representation of G on this subspace, hence a linear action on kn making the morphism β : X � kn given by β(x) = (h1(x), . . . , hn(x)) into a G-morphism. Since G is geometrically reductive there is an invariant polynomial f that has the value zero on β(W1 ) and the value 1 on β(W2). f ∩ β is the desired polynomial. � � i The main theorem for quotients of reductive group actions on aﬃne varieties is the following: 17 Theorem 6.2. Let G be a reductive group acting on an aﬃne variety X. Then there exists a quotient aﬃne variety Y and a G-invariant, surjective morphism ζ : X � Y such that (1) For any open set U ≥ Y , the ring homomorphism ζ� : A(U ) � A(ζ−1 (U )) is an isomorphism of A(U ) with A(ζ−1(U ))G . (2) If W ≥ X is a closed invariant subset, then ζ(W ) is closed in Y . (3) If W1 and W2 are disjoint closed invariant sets, then their images under ζ are disjoint. Proof. The main technical results are provided by a theorem of Haboush and a theorem of Nagata. Theorem 6.3 (Haboush). Any reductive group G is geometrically reductive. Theorem 6.4 (Nagata). Let G be a geometrically reductive group acting rationally on a ﬁnitely generated k-algebra R. Then the ring of	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
ductive group acting rationally on a ﬁnitely generated k-algebra R. Then the ring of invariants RG is ﬁnitely generated. In view of these theorems A(X)G is ﬁnitely generated. Hence we can let Y = Spec A(X)G . The inclusion of A(X)G � A(X) induces a morphism ζ : X � Y . � The claimed properties are easy to check for ζ. Remark 6.5. The following are straightforward observations: (1) For any open subset U ≥ Y , (U, ζ) is a categorical quotient of ζ−1(U ) by G. (2) The images of two points in X coincide if and only if the orbit closures of these two points intersect. Consequently, Y will be an orbit space if and only if the orbits of the G action on X are closed. Remark 6.6. We will not prove Haboush’s theorem here. The interested reader may consult the original paper [Hab]. Over the complex numbers reductive, ge ometrically reductive and linearly reductive coincide. This follows from the fact that any ﬁnite dimensional representation is decomposible to irreducible represen tations. Projection to the one-dimensional invariant subspace produces the desired invariant linear functional. We now sketch the proof of Nagata’s theorem. Since R is a ﬁnitely generated k-algebra, we can pick generators f1, dots, fn that generate R. We can also assume that the subspace spanned by the fi is G-invariant. (If not, we can replace it by a minimal G-invariant subspace, which is �	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
If not, we can replace it by a minimal G-invariant subspace, which is ﬁnite-dimensional by the argument in Lemma 6.1.) We thus obtain a linear G action on the subspace spanned by fi by setting Let S = k[X1, . . . , Xn]. There is an action of G on S by setting f g = i � j �i,j (g)fj . X g = i � j �i,j (g)Xj . There is a k-algebra homomorphism from S to R sending Xi to fi that is compatible with the G actions. We are thus reduced to proving Nagata’s theorem in the case 18 when G acts on S preserving degree, Q ≥ S is a G-invariant ideal with the induced action on R = S/Q. Under these assumptions we would like to see RG is ﬁnitely generated. Suppose not. Since S is Noetherian, there exists an ideal Q maximal among those that are G-invariant such that RG where R = S/Q is not ﬁnitely generated. Then if J �= 0 is a G-invariant homogeneous ideal in R, then (R/J)G is ﬁnitely generated. Suppose ﬁrst there is a homogeneous ideal Q with the desired properties. I claim that (R/J)G is integral over RG/(J ≤ RG). Suppose f ⊗ (R/J)G . Pick h ⊗ R such that the image of h in R/J is f . We would like to ﬁnd h0 ⊗ RG such that (h)t − h0 for some positive integer t is in RG . Look at the ﬁnite-dimensional, G-invariant subsapce M generated by hg . [Unfortunately, there is potential for confusion between hg and (h)t . The ﬁrst denotes the g-translate of h	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
for confusion between hg and (h)t . The ﬁrst denotes the g-translate of h, the second denotes the t-th power of h. To distinguish between these two, we will put paren theses around h in the latter case.] Since J is invariant, hg − h is in J for every g. We conclude that M ≤ J has codimension 1 in M . We can write every element in M uniquely as ah + h≤ where a ⊗ k and h≤ ⊗ M ≤ J. Sending ah + h≤ to a deﬁnes a G-invariant linear functional l on M . There is an action of G also on M � . If we let h, j2, . . . , jn be a basis of M where ji ⊗ M ≤ J, we can identify M � with kr in terms of the dual basis. The linear functional l corresponds to the vector (1, 0, . . . , 0). Since G is geometrically reductive, there exists an invariant homogeneous polynomial F ⊗ k[X1, . . . , Xn] of t does not vanish. Consider the morphism degree t → 1 such that the coeﬃcient of X1 k[X1, . . . , Xn] sending X1 to h and Xi to ji for i > 1.If h0 is the image of F , ht − h0 belongs to J. We conclude that (R/J)G is integral over RG/(J ≤ RG). If A is a ﬁnitely generated k-algebra which is integral over a subalgebra B, then B is ﬁnitely generated. Hence in our case, RG/(J ≤ RG) is ﬁnitely generated. In fact, (R/J)G is a ﬁnite RG/(J ≤ RG)-module. Choose a non-zero homogeneous element f of RG of degree	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
ﬁnite RG/(J ≤ RG)-module. Choose a non-zero homogeneous element f of RG of degree at least one. If f is not a zero-divisor, f R ≤ RG = f RG . Since RG/f RG is ﬁnitely generated, (RG/f RG)+ is ﬁnitely generated as an ideal. Hence RG is ﬁnitely generated as an + ideal in RG . Hence RG is a ﬁnitely generated k-algebra. Exercise 6.7. Modify the last paragraph of the proof in case f is a zero-divisor. Hint: Consider the homogeneous ideal I of elements of R that annihilate f . Since RG/(f R ≤ RG) and RG/I ≤ RG are both ﬁnitely generated, there is a ﬁnitely generated subalgebra of RG that surjects onto both these algebras In order to handle the non-homogeneous case, we may assume that RG is a domain. By the homogeneous case S G is ﬁnitely generated. RG is integral over SG/Q≤SG . It suﬃces to show that the ﬁeld of fractions of RG is a ﬁnitely generated extension of k. Let T be the set of non-zero divisors of R. Form the ring of fractions of R with respect to T . Let m be the maximal ideal. The ﬁeld of fractions of RG may be identiﬁed with a subﬁeld of T −1R/m. Since T −1R/m is the ﬁeld of fractions of the ﬁnitely generated k-algebra R/m ≤ R, this follows. Example 6.8. Everyone’s favorite example is the action of GL(n) on the space of n × n matrices Mn by conjugation. The space of matrices is isomorphic to aﬃne	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
matrices Mn by conjugation. The space of matrices is isomorphic to aﬃne space An . Hence, the coordinate ring is k[ai,j ], 1 ∼ i, j ∼ n. Any conjugacy class 19 2 has a representative in Jordan canonical form which is unique upto a permutation of the Jordan blocks. Since the set of eigenvalues of a matrix is invariant under conjugation, we see that the elementary symmetric polynomials of the eigenval ues, i.e. the coeﬃcients of the characteristic polynomial, are invariant under the action. Conversely, suppose that a polynomial is invariant under conjugation. If the eigenvalues are distinct, we can diagonalize the matrix by connjugation. Hence the polynomial must be a symmetric function of the eigenvalues. If the eigenvalues are repeated, the diagonal matrix is in the closure of the orbits with non-trivial Jordan blocks. We conclude that any invariant polynomial is a symmetric polyno mial of the eigenvalues. Since the elementary symmetric polynomials generate the ring of symmetric polynomials, we conclude that the ring of invariant functions is generated by the coeﬃcients of the characteristic polynomial. Now we would like to extend the discussion from actions of reductive groups on aﬃne varieties to actions on projective varieties. Suppose we have a group acting on a projective variety X ≥ Pn . A linearization of the action of G is a linear action on kn+1 which induces the given action on X. More generally, let X be a variety, G a group acting on it and L a line bundle on X. A linearization of the action of G with respect to L is a linear action on L that induces the action of G on X. Deﬁnition 6.9. A point x ⊗ X is called semi	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
on X. Deﬁnition 6.9. A point x ⊗ X is called semi-stable if there exists an invariant homogeneous polynomial that does not vanish on x. A point x ⊗ X is called stable if there exists an invariant polynomial f that does not vanish on x, the action of G on Xf is closed and the dimension of the orbit of x is equal to the dimension of G. These depend not only on the action, but the chosen linearization. Denote the locus of semi-stable points by X ss and the locus of stable points by X s . Remark 6.10. Note that the semi-stable points are precisely those that do not contain 0 in the closure of their orbits. Both X ss and X s are clearly open (possibly empty) in X. The main theorem of G.I.T. is the existence of a good quotient of the semi-stable locus whose restriction to the stable locus is a geometric quotient. We will call a quotient a good quotient if it satisﬁes the conditions of Theorem 6.2. We will call a good quotient that is also an orbit space a geometric quotient. Theorem 6.11. Let X be a projective variety in Pn . Then for every linear action of a reductive group G on X (1) There exists a good quotient (Y, ζ) of X ss by G and Y is projective. (2) There exists an open subset Y s of Y such that ζ−1(Y s) = X s and (Y s, ζ) is a geometric quotient of X s . In view of this theorem it is important to determine the stable and semi-stable loci for reductive group actions on projective varieties. Unfortunately, this in gen	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
is important to determine the stable and semi-stable loci for reductive group actions on projective varieties. Unfortunately, this in gen eral is a very diﬃcult problem. There is one instance where stability and semi- stability is easy to determine. Deﬁnition 6.12. A one-parameter subgroup is a homomorphism � : Gm � G. Any action of k� on kn+1 can be diagonalized. Hence, there exists a basis e0, . . . , en such that the action of the one-parameter subgroup � is given by �(t)ei = 20 tri ei for some integers ri. If ˆ = x xiei, then Deﬁne � x�(t)ˆ = tri xiei. � i µ(x, �) = max{−ri \| xi �= 0 }. Theorem 6.13 (The Hilbert-Mumford criterion of stability). Let G be a reductive group acting linearly on a projective variety X ≥ Pn . Then: (1) x is semi-stable if and only if for every one-parameter subgroup � of G µ(x, �) → 0. (2) x is stable if and only if for every one-parameter subgroup � of G µ(x, �) > 0. Proof. The challenging part of the theorem is to produce a one-parameter subgroup that has the wrong µ invariant if x is not semi-stable. We will sketch Hilbert’s proof for the case G = SL(m). The general case follows the same general line of argument (see §2.1 [FKM]). g / x where ˆ Let K be the ﬁeld of fractions of R = k[[T ]]. If x is not stable, then the morphism G � kn+1 given by sending g to gˆ x is a lift of x is not	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
1 given by sending g to gˆ x is a lift of x is not proper. By the g ⊗ SL(m, K) such that ¯x ⊗ Rn+1 , valuative criterion of properness, there exists ¯ but ¯ ⊗ SL(m, R). We can, however, clear denominators so that T rg¯ ⊗ SL(m, R) for some r. The ring R is a P.I.D., hence we can decompose ¯ = g¯1dg¯2 where g1 and g2 are in SL(m, R) and d is a diagonal matrix consisting of entries T w1 , . . . , T wm for some integers wi whose sum is zero (since the resulting matrix has to be in SL(m, K). This is the point in the proof where we are using that G = SL(m). To prove the theorem for general groups one needs to use a theorem of Iwahori which asserts that the double coset in G(R)\G(K)/G(R) for a reductive group can be represented by a one-parameter subgroup. gˆ g Let g2 be the matrix obtained by setting T = 0 in g¯2. The de-stabilizing one- parameter subgroup is deﬁned by �(t) = g −1diag(tw1 , . . . , twm )g2. 2 Diagonalize the action of � on kn+1 with respect to a basis e0, . . . , en as above. We would like to show that if ˆ = 0, then the weight ri of the action on ei is non-negative. We can also consider the basis e0, . . . , en as a basis of K n+1 . Then g −1dg2ei = T ri ei. In particular,	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
as a basis of K n+1 . Then g −1dg2ei = T ri ei. In particular, 2 −1 g¯ = g −1 g¯1dg¯2 = (g −1dg2)g xi � g¯1 −1 g 2 Therefore, the i-th component of g −1 g¯1 −1 g 2 in R, we conclude that ri → 0. −1 2 g¯1 2 ¯ x. Consequently, the i-th component of g −1 g2 ˆ g2 ˆ 2 gˆ g¯2. −1 ¯x is T ri times the i-th component of ¯ x is in T −ri R. Since it is also � 2 −1 2 Exercise 6.14. Modify the previous argument to obtain the theorem for the semi- stable case. Example 6.15 (Points on P1). Consider the action of SL(2) on the homogeneous polynomials of degree d in two variables. Let � be a one-parameter subgroup of SL(2). If we diagonalize the action of � on k2 by diag(ta, t−a) in coordinates (x, y), then the monomials xiyd−i diagonalizes the action of � on homogeneous 21 polynomials of degree d. The weight of the action on xiyd−i is a(2i − d). If we want the weight to be negative, then the coeﬃcient of one monomial xiy with 2i − d < 0 has to be non-zero. This means that a homogeneous polynomial is stable if and only if it does not have any zeros with multiplicity → d/2. Similarly, a homogeneous polynomial	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
is stable if and only if it does not have any zeros with multiplicity → d/2. Similarly, a homogeneous polynomial is semi-stable if and only if it does not have any zeros with multiplicity > d/2. d−i i j Example 6.16 (Cubic plane curves). Consider the action of SL(3) on the homo geneous polynomials of degree 3 in three variables. If we diagonalize the action of a one-parameter subgroup � in terms of the coordinates x1, x2, x3 such that �(t)xi = twi xi, then the basis given by monomials x1x2x 3−i−j diagonalizes the action of � on degree 3 homogeneous polynomials. The weight of the action on j 3−i−j i x2x is given by iw1 + jw2 + (3 − i − j)w3. We can visualize the one param x1 3 eter subgroup in terms of barycentric coordinates. The one-parameter subgroups correspond in this picture to lines pivoted around the point (i, j, 3−i−j) = (1, 1, 1). If we move the line without crossing any integral points on the triangle, we do not change the conditions for stability. Also the picture is invariant under the sym metries of the triangle. Analyzing the coeﬃcients we see that a cubic is stable if and only if it is smooth. Similarly a cubic is semi-stable if and only if it has ordi nary double points. Note that the G.I.T. quotient of the stable locus in this case constructs the j-line. 3 Exercise 6.17. Try to generalize	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
the stable locus in this case constructs the j-line. 3 Exercise 6.17. Try to generalize the previous example to the action of SL(3) on homogeneous polynomials of degree 4, 5, 6, .... In particular, describe what kinds of singularities are allowed on stable curves of degree 4, 5, 6... 6.2. The construction of M g . In view of Theorem 6.11 in order to construct M g we need to show that the N -canonically embedded Deligne-Mumford stable curves are stable points for the SL(n + 1)-action on the Hilbert scheme and that they form a closed subset. The details of this veriﬁcation are involved. You may ﬁnd good accounts in [HM] and [Mum3]. We would like to apply the Hilbert-Mumford criterion to the action of SL(n + 1) on HilbP (m)(Pn). Fix a one-parameter subgroup � of SL(n + 1). Suppose in terms of homogeneous coordinates xi that diagonalize the action, the weights are w0, . . . , wn. Of course, as usual we have that i wi = 0. Recall that we exhibited the Hilbert scheme as a subscheme of the Grassmannian G(P (m), H 0(Pn , OPn (m))) for m greater than or equal to the regularity of all the ideal sheaves with Hilbert polynomial P . The Grassmannian has natural Pl¨ucker coordinates consisting of P (m)-element subsets of monomials in the xi of degree m. This basis also diag P (m) H 0(Pn , OPn (m)). The weight on the onalizes the action of SL(n + 1) on Pl¨ ucker coordinate { Yj1 , . . . , YjP (m) } where Y	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
(n + 1) on Pl¨ ucker coordinate { Yj1 , . . . , YjP (m) } where Yji = is given by mji ,r xr r � � wr mji ,r. � i,r The Hilbert-Mumford criterion for semi-stability then translates to the condition that for each one parameter subgroup, there should be a non-vanishing Pl¨ucker coordinate whose weight is non-positive. We begin by showing that the m-th Hilbert points of smooth, non-degenerate curves embedded by a complete linear series of degree d → 2g are stable for the SL(n + 1) action. 22 Theorem 6.18 (Stability for smooth curves). Let C be a smooth curve of genus g → 2 embedded in projective space Pd−g by a complete linear system of degree d at least 2g. Then C is Hilbert stable. Moreover, there exists M such that for all m → M , the m-th Hilbert point of non-degenerate, smooth curves of degree d and genus g in Pd−g is stable. Sketch. The proof is an application of the Hilbert-Mumford criterion. � Deﬁnition 6.19 (Potential stability). A connected curve C of degree d and genus g in Pd−g+1 is called potentially stable if (1) The embedded curve C is non-degenerate. (2) The abstract curve C is moduli semi-stable. (3) The linear series embedding C is complete and non-special (i.e. has h1 = 0). (4) If C ≤ is a complete subcurve of C of	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
has h1 = 0). (4) If C ≤ is a complete subcurve of C of arithmetic genus g≤ meeting the rest of the curve C in k points, then the following estimate holds degC (OC (1)) − d g − 1 k (gC − 1 + ) 2 k . 2 ∼ � � � � � � � � Remark 6.20. Observe that if C ≤ is a smooth rational curve meeting the rest of the curve in exactly two points (k = 2), then the term gC − 1 + k/2 = 0, hence the degree of C ≤ has to be 1. In other words, C ≤ is a line. By the same argument, if C ≤ is a nodal tree of smooth rational curves meeting the rest of C in exactly two points, then C ≤ is a smooth rational curve since the degree is at most one. Furthermore, C ≤ cannot meet the rest of the curve in only one point. Recall that �C\|C is the dualizing sheaf �C twisted by the nodes connecting C ≤ to C. Hence, deg(�C\|C ) = 2gC − 2 + k. Condition (4) has the following alternative useful expression deg C ≤ − d � � � � deg(�C\|C ) deg(�C ) � � � � ∼ k . 2 Theorem 6.21 (Potential stability). Let g → 2 and d > 9(g − 1). Then there is an integer M depending only on d and g such that if m → M and C ⊗ Pd−g is a connected curve with semi-stable m-th Hilbert point, then C is potentially	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
−g is a connected curve with semi-stable m-th Hilbert point, then C is potentially stable. The proof of this theorem is quite lengthy eventhough the strategy is straight forward. We suppose C has a geometric property that violates potential stability. Under this assumption we construct a one-parameter subgroup that destabilizes the Hilbert point of C contradicting the assumption that the m-th Hilbert point of C was semi-stable. We ﬁrst assume Theorem 6.21 and deduce from it the existence of the coarse moduli space M g . Fix an integer r → 5. Consider r-canonically embedded stable ∗r is very ample for r → 3, every Deligne-Mumford stable curve curves. Since �C has a representative in the Hilbert scheme H = Hilbr(2g−2)+1−g(Pr(2g−2)−g ). Now consider the subscheme H of H subscheme of the Hilbert scheme parameterizing r-canonically embedded Deligne-Mumford stable curves. Let H ss denote the inter- section of H with the semi-stable locus of H. Since r → 5, we have that the degree of the curves are at least 10(g − 1) > 9(g − 1). Therefore, the assumption of the Potential Stability Theorem is satisﬁed. We conclude that every semi-stable point ˆof H is potentially stable. ˆ ˆ ˆ 23 Lemma 6.22. The locus H ss is closed in semi-stable locus of the Hilbert scheme H ssˆ . Proof. To show that H ss is closed we need to show that the inclusion H ss � ˆ	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
H ssˆ . Proof. To show that H ss is closed we need to show that the inclusion H ss � ˆ H ss is proper. By the valuative criterion of properness it suﬃces to check that given a map from the spectrum of a DVR to ˆH ss whose generic point lies in H ss, the closed point also lies in H ss . Given such a map consider the universal curve CR over Spec (R). There are two line bundles on CR, the relative dualizing sheaf �CR/R and OCR (1). These two are isomorphic except possibly at the central ﬁber. To conclude the lemma we need to show that they also agree on the central ﬁber. Hence the i aiCi is a linear combination of the central two diﬀer by OCR (− ﬁber. We need that ai = 0 for all i. We can assume that ai → 0 for all i with at ≤ be the subcurve of the central ﬁber D where ai > 0 and least one ai = 0. Let C1 C ≤ be the subcurve of the central ﬁber D where ai = 0. We see that all ai = 0 as follows. A local equation of OCR (− i aiCi) is identically zero on every component ≤ . In particular, the local equation vanishes at the ≤ and on no component of C1 of C2 k points of intersection between C1 ≤ . We then have that � ≤ and C2 i aiCi) where � � 2 k ∼ degD (OCR (− � i aiCi) ∼ degD (OCR (1)) − degC (OCR (1)\|C ) 2 degC 2 (	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
degD (OCR (1)) − degC (OCR (1)\|C ) 2 degC 2 (�\|C 2 ) 2 degD (�\|C 2 ) ∼ k . 2 � Lemma 6.23. Every curve C whose Hilbert point lies in H ss is Deligne-Mumford stable. Proof. By the potential stability theorem C is semi-stable. In order to show that it is stable we need to check that there are no rational curves that intersect the rest of the curve in only two points. On a rational curve meeting the rest of C in two points, the degree of the dualizing sheaf of C is zero whereas OC (1) is very ample. Since these two coincide for points in H ss, we conclude that C must be � Deligne-Mumford stable. Lemma 6.24. Every Deligne-Mumford stable curve of genus g has a model in H ss . Proof. Every moduli stable curve C is embedded in Pr(2g−2)−g by its � ∗r . We need C to show that the Hilbert point of C lies in H ss . If C is smooth, we already know this by Theorem 6.18. To deduce it for singular Deligne-Mumford stable curves, we take a one-parameter deformation of C to a smooth curve of genus g over the spectrum of a DVR R. If we embed this curve r-canonically, we get a map from Spec R to the Hilbert scheme. The generic point lies in H ss . Since the G.I.T. quotient of the Hilbert scheme ˆH ss by the action of the special linear group is projective, after a	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
scheme ˆH ss by the action of the special linear group is projective, after a base change we can extend the map to ˆH ss . Since H ss is closed, the image of the map lies in H ss . Pulling back the universal curve we obtain a semi-stable reduction of a family of stable curves. By the uniqueness of semi-stable reduction this family has to agree with our original family. Since the curves H ss are actually stable, the central ﬁber of both families have to be projectively equivalent. The � lemma follows. Lemma 6.25. Every curve whose Hilbert point lies in H ss is Hilbert stable. Proof. We need to show that every point in H ss has closed orbit and the stabilizer of a point in H ss is ﬁnite. Suppose the stabilizer is not ﬁnite, then the curve 24 C would have inﬁnitely many automorphisms contradicting that Deligne-Mumford stable curves have only ﬁnitely many automorphisms. If the orbit is not closed, then the closure would contain a semi-stable orbit with positive dimensional stabilizer. � Again we would obtain a contradiction. Lemma 6.26. The locus H ss is non-singular. C proper and ﬂat over Spec k[[t1, . . . , tr ]] where r = dim Ext1(�1 Proof. Recall that given a Deligne-Mumford stable curve C, there exists a formal scheme ˜ C , OC ) such that the special ﬁber is isomorphic to C. Moreover, for a stable curve the versal deformation is universal and algebrizable and the generic ﬁber is smooth. ˜ Let [C] ⊗ H ss be a point. Let	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
gebrizable and the generic ﬁber is smooth. ˜ Let [C] ⊗ H ss be a point. Let C be the universal formal deformation of C over B = Spec k[[t1, . . . , tr]]. Set S be the formal completion of H ss at [C]. By the universal property of the Hilbert scheme we get a map S � H ss . By the universal property there exists a unique morphism f : S � B such that the pull-back of the universal curve is S ×B C. The Lemma follows from the claim that f : S � B is � formally smooth. ˜ One important aspect of the G.I.T. construction is that the projectivity of M g is immediate. Another important consequence is the irreducibility of the moduli space of curves over an algebraically closed ﬁeld of any characteristic. Originally Deligne and Mumford developed the theory of Deligne-Mumford stacks to prove the irreducibility in all characteristics and for all genus in [DM]. Theorem 6.27. The moduli space M g is projective. Theorem 6.28. The moduli space M g is irreducible (and reduced) over any alge braically closed ﬁeld. Proof. Soon we will see that the moduli space of curves in characteristic zero is irreducible. There are many ways of seeing this. We will use Teichm¨uller theory to construct Mg as the quotient of a bounded, contractible domain in C3g−3 . Al ternatively, one can exhibit every smooth curves as a branched cover of P1 . When the number of branch points is large relative	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
exhibit every smooth curves as a branched cover of P1 . When the number of branch points is large relative to the degree of the map, using the combinatorics of the symmetric group one may show that the space of branched covers of P1 is irreducible. Suppose now that the characteristic of the ﬁeld k is positive. Let R be a discrete valuation ring whose quotient ﬁeld has characteris tic zero and whose residue ﬁeld is k. The construction outlined so far works over R /PGL � Spec R is connnected, by Zariski’s Spec R. Since the generic ﬁber of H ss R /PGL ∗ k is connected. Since this is an orbit space H ss connectedness theorem H ss is connected. Since it is smooth, it is reduced and irreducible. Consequently M g is also irreducible. M g is also reduced because the structure sheaf of the quotient is � the sheaf of invariants of the structure sheaf of H ss . k Finally we enumerate the steps that one carries out in order to prove the Poten tial Stability Theorem. We assume that a geometric condition violating potential stability occurs on a curve. We then produce a one-parameter subgroup destabiliz ing that point, hence showing that it is not a Hilbert stable point. Unfortunately the number of cases and calculations needed to give a complete proof is rather large. Since we will not use these techniques later in the course, we will just sketch a few sample cases. A complete proof can be found on pages 35-87 of [G]. 25 Claim 6.29. The ﬁrst claim is that if a curve C is Hilbert stable, then Cred is not contained in a hyperplane. If the curve is degenerate, then the map H 0(OPn (1)) � H	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
contained in a hyperplane. If the curve is degenerate, then the map H 0(OPn (1)) � H 0(Cred, OCred (1)) has non-trivial kernel. Use the ﬁltration that assigns weight −1 to sec tions vanishing on Cred and weight w > 0 to the others so that the average weight is 0. There exists an integer q such that the q-th power of the ideal sheaf of nilpotents in OC is zero. Hence no monomial that contains more than q factors of weight −1 can be zero. Provided we choose m such that (m − q)w > q, every element of a monomial basis of H 0(C, OC (m)) has positive weight. Hence, C is not Hilbert semi-stable. From now on we may assume that the linear span of our curves in Pn . This argument is the blueprint for the other arguments. We will give very few details for the other ones. Claim 6.30. The second claim is that every component of C is generically reduced. Claim 6.31. The third claim is that every singularity of Cred is a double point. If p is a point of multiplicity 3 or more, the two-step ﬁltration assigning weight 0 to the sections vanishing at p and weight one to the others is destabilizing. Claim 6.32. Every double point of Cred is a node. Claim 6.33. H 1(Cred, OC (1)) = 0 Claim 6.34. C is reduced. From these claims it follows that the ﬁrst three conditions of the deﬁnition of potential stability hold. The ﬁnal step is to show that the estimate in (4	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
potential stability hold. The ﬁnal step is to show that the estimate in (4) holds. This is done by showing that if not the ﬁltration FC is destabilizing. References [Ab] [DM] [Ed] [Fan] [G] [Gr] S. S. Abhyankar. Resolution of singularities of arithmetical surfaces. In Arithmetical Algebraic Geometry (Proc. Conf. Purdue Univ., 1963), pages 111–152. Harper & Row, New York, 1965. P. Deligne and D. Mumford. The irreducibility of the space of curves of given genus. IHES Publ. Math. 36(1969), 75–110. D. Edidin. Notes on the construction of the moduli space of curves. In Recent progress in intersection theory (Bologna, 1997), Trends Math., pages 85–113. Birkh¨auser Boston, Boston, MA, 2000. B. Fantechi. Stacks for everybody. In European Congress of Mathematics, Vol. I (Barcelona, 2000), volume 201 of Progr. Math., pages 349–359. Birkh¨auser, Basel, 2001. D. Gieseker. Lectures on moduli of curves, volume 69 of Tata Institute of Fundamen tal Research Lectures on Mathematics and Physics. Published for the Tata Institute of Fundamental Research, Bombay, 1982. A. Grothendieck. Techniques de construction et th´ emes d’existence en g´ emas de Hilbert. In S´ alg´ 221,	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
emes d’existence en g´ emas de Hilbert. In S´ alg´ 221, 249–276. Soc. Math. France, Paris, 1995. etrie eminaire Bourbaki, Vol. 6, pages Exp. No. ebrique. IV. Les sch´ eom´ eor` [Hab] W. J. Haboush. Reductive groups are geometrically reductive. Ann. of Math. (2) [HM] [Ha] [KM] [K] 102(1975), 67–83. J. Harris and I. Morrison. Moduli of curves. Springer-Verlag, 1998. R. Hartshorne. Algebraic geometry. Springer-Verlag, New York, 1977. Graduate Texts in Mathematics, No. 52. S. Keel and S. Mori. Quotients by groupoids. Ann. of Math. (2) 145(1997), 193–213. J. Koll´ar. Rational curves on algebraic varieties, volume 32 of Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. A Series of Modern Surveys in Mathematics. Springer- Verlag, Berlin, 1996. 26 [LM-B] G. Laumon and L. Moret-Bailly. Champs alg´ebriques, volume 39 of Ergebnisse der Math ematik und ihrer Grenzgebiete. 3. Folge. A Series of Modern Surveys in Mathematics [Results in Mathematics and Related Areas. 3rd Series. A Series of Modern Surveys in Mathematics]. Springer-Verlag, Berlin, 2000. M. Lieblich. Groupoids and quotients in algebraic geometry. In Snowbird lectures in algebraic geometry, volume 388 of Contemp	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
algebraic geometry. In Snowbird lectures in algebraic geometry, volume 388 of Contemp. Math., pages 119–136. Amer. Math. Soc., Providence, RI, 2005. [Li] [Mum1] D. Mumford. Further pathologies in algebraic geometry. Amer. J. Math. 84(1962), 642– 648. [Mum2] D. Mumford. Lectures on curves on an algebraic surface. With a section by G. M. Bergman. Annals of Mathematics Studies, No. 59. Princeton University Press, Princeton, N.J., 1966. [Mum3] D. Mumford. Stability of projective varieties. Enseignement Math. (2) 23(1977), 39–110. [FKM] D. Mumford, J. Fogarty, and F. Kirwan. Geometric invariant theory, volume 34. Springer- [Ne] [Se] Verlag, Berlin, third edition, 1994. P. E. Newstead. Introduction to moduli problems and orbit spaces, volume 51 of Tata Institute of Fundamental Research Lectures on Mathematics and Physics. Tata Institute of Fundamental Research, Bombay, 1978. E. Sernesi. Topics on families of projective schemes, volume 73 of Queen’s Papers in Pure and Applied Mathematics. Queen’s University, Kingston, ON, 1986. 27	https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf
MIT OpenCourseWare http://ocw.mit.edu 6.641 Electromagnetic Fields, Forces, and Motion, Spring 2005 Please use the following citation format: Markus Zahn, 6.641 Electromagnetic Fields, Forces, and Motion, Spring 2005. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7: Polarization and Conduction I. Experimental Observation A. Fixed Voltage - Switch Closed (v = V o ) As an insulating material enters a free-space capacitor at constant voltage more charge flows onto the electrodes; i.e. as x increases, i increases. B. Fixed Charge - Switch open (i=0) As an insulating material enters a free space capacitor at constant charge, the voltage decreases; i.e. as x increases, v decreases. II. Dipole Model of Polarization A. Polarization Vector P = N p = N q d ( p = q d dipole moment) N dipoles/Volume ( P is dipole density) d + q −q 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 1 of 27 Courtesy of Krieger Publishing. Used with permission. 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 2 of 27 Cite as: Markus Zahn, course materials for 6.641 Electromagnetic Fields, Forces, and Motion, Spring 2005. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Q inside V = − (cid:118)∫ q N d S i da = ∫ ρP	https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf
]. Q inside V = − (cid:118)∫ q N d S i da = ∫ ρP dV V paired charge or equivalently polarization charge density Qinside V = − P i da = − ∇ i P dV = (cid:118)∫ S ∫ V ρ dV P ∫ V (Divergence Theorem) P = q N d ∇ i P = −ρP B. Gauss’ Law ∇ i (ε E) = ρ o total = ρ + ρ = ρ − ∇ i P u P u unpaired charge density; also called free charge density ∇ i (ε o + ) = ρ E P u D = εo E + P Displacement Flux Density ∇ i D = ρu C. Boundary Conditions 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 3 of 27 Da − Db ⎤ ∇ i D = ρu ⇒ (cid:118)∫ D i da = ∫ ρu dV ⇒ n i ⎡ ⎣ ⎦ = σ su S V P a − Pb ⎤ ∇ i P = −ρ P ⇒ (cid:118)∫ P i da = −∫ ρP dV ⇒ n i ⎡ ⎣ ⎦ = −σ sp S V ∇ i (ε E) = ρ + ρ ⇒ (cid:118)∫ ε E i da = ∫ (ρ + ρ ) dV ⇒ n i ε ⎣ u u o o o P P ⎡Ea − Eb ⎦ ⎤ = σ + σ su sp S V D. Polarization Current Density ∆Q = q N dV = q N d da	https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf
su sp S V D. Polarization Current Density ∆Q = q N dV = q N d da i = P i da [Amount of Charge passing through surface area element da ] d ip = ∂∆Q ∂t = ∂P ∂t i da = Jp i da polarization current density [Current passing through surface area element da ] Jp = ∂P ∂t Ampere’s law: ∇ x H = Ju + Jp + εo ∂E ∂t = Ju + ∂P ∂t + εo ∂E ∂t = Ju + ∂ (ε E + P) ∂t o = Ju + ∂ D ∂t 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 4 of 27 III. Equipotential Sphere in a Uniform Electric Field ) lim Φ r, θ = − E r cos θ r → ∞ ( o ⎡Φ = − E z = −E r cos θ⎤ ⎣ ⎦ o o Φ (r = R, θ) = 0 Φ (r, θ) = − Eo ⎡ ⎢r − ⎣ 3 ⎤ R 2 r ⎦ ⎥ cos θ This solution is composed of the superposition of a uniform electric field plus the field due to a point electric dipole at the center of the sphere: Φ dipole = p cos θ 4πε ro 2 with p = 4 πε E o o 3 R This dipole is due to the surface charge distribution on the sphere. σ (r = R, s θ) = ε E (r = R, o r θ) = − ε ∂Φ o �	https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf
θ) = ε E (r = R, o r θ) = − ε ∂Φ o ∂r r R = = ε oEo ⎡ ⎢1 + ⎣ 3 2R 3 r ⎤ ⎥ cos r R ⎦= θ = εo o 3 E cos θ 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 5 of 27 IV. Artificial Dielectric E = v d , σ = ε E = s ε v d q = σsA = ε A d v C = q v = ε A d E d ε _ υ + Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. For spherical array of non-interacting spheres (s >> R) P = 4 π ε R 3 E i z ⇒ P = N p = 4 π ε R 3 E N o o o z o z _ N = 1 3 s ⎡ ⎛ P = εo ⎢4 π ⎜ ⎝ ⎢ ⎣ R s 3 ⎤ ⎞ ⎟ ⎥ E = ψ e εo E ⎠ ⎥ ⎦ ⎛ ⎛ ⎜ ψ e = 4 π ⎜ ⎜ ⎝ ⎝ 3 ⎞ R ⎞ ⎟ ⎟ ⎟ s ⎠ ⎠ ψe (electric susceptibility) D = εo E + P = ε o ⎡⎣1 + ψ e ⎤⎦ E = ε E εr (relative dielectric constant) ε ε= r ε o ε = ⎡1 + ψ ⎤ = ε e	https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf
lectric constant) ε ε= r ε o ε = ⎡1 + ψ ⎤ = ε e ⎦ o ⎣ ⎛ ⎛ ⎜1 + π 4 ⎜ o ⎜ ⎝ ⎝ R s 3 ⎞ ⎞ ⎟ ⎟ ⎟ ⎠ ⎠ 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 6 of 27 V. Demonstration: Artificial Dielectric Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 7 of 27 Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. E = ⇒ σ = ε E = s v d ε v d q = σsA = ε A d v ⇒ C = = q v ε A d ∆i = ω ∆C V = v o R s ∆C = ( − o ) A ε ε d ⎛ = 4 π εo ⎜ ⎝ ⎞R ⎟ ⎠s 3 A d R=1.87 cm, s=8 cm, A= (0.4)2 m2, d=0.15m ω =2π(250 Hz), Rs=100 k Ω , V=566 volts peak ∆ C=1.5 pf v = ω ∆C R V s 0 =(2π) (250) (1.5 x 10-12) (105) 566 = 0.135 volts peak 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 8 of 27 VI. Plasma Conduction Model (Classical) m dv + + dt	https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf
x 10 − 12 farads/m ω = p− 2q n − m − ε − ≈ 5.6 x 10 11 rad/s fp − = ωp − ≈ 9 x 10 2π 10 Hz 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 9 of 27 B. Drift-Diffusion Conduction [Neglect inertia] 0 m + +dv dt = q E + − m ν v − + + + + ∇ (n k T ) n+ ⇒ v = + q m +ν + + E − k T m + ν +n+ ∇n + 0 m − dv − dt = −q E − J = q + + n v+ = + − m ν v − − − − ) ( − ∇ n k T n− ⇒ v = − −q− E − m −ν − k T m −ν −n− ∇n − 2 + q n+ m +ν + E − + q k T m +ν + ∇n+ J = − −q− n− v− = − 2q n m −ν − − E + − q k T m −ν − ∇n− ρ + = q+ n+ , ρ − = −q− n− J = ρ µ E − D ∇ρ + + + + + J = − −ρ − µ − E − D −∇ρ − µ + = q+ m +ν + µ − = q− m −ν − , , D + = k T m + ν + D − = k T m − ν − charge mobilities Molecular Diffusion Coefficients D +	https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf
+ D − = k T m − ν − charge mobilities Molecular Diffusion Coefficients D + = µ + D − = µ − k T = thermal voltage (25 mV @ T ≈ 300o K) q Einstein’s Relation 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 10 of 27 C. Drift-Diffusion Conduction Equilibrium (J+ = J− = 0) J = 0 = ρ µ E − D ∇ρ = −ρ µ ∇Φ − D ∇ρ + + + + + + + + + J = 0 = −ρ µ E − D ∇ρ = − ρ µ ∇Φ − D ∇ρ − − − − − − − − ∇Φ = − D + ∇ρ + = ρ µ + + −k T q ∇ (ln ρ + ) ∇Φ = D − ρ µ − − ∇ρ − = k T q ∇ (ln ρ − ) ρ + = ρoe q / kT − Φ Boltzmann Distributions ρ = −ρ e q / kT o + Φ − ρ Φ( + = 0 ) = ρ Φ = 0 ) = ρ ( − − o [Potential is zero when system is charge neutral] 2 ∇ Φ = −ρ ε = − + (ρ + ρ− ) ε = −ρo ε ⎣ q / kT ⎡e− Φ − e + Φ q / kT ⎤ = ⎦ 2ρo ε sinh q Φ kT (Poisson-Boltzmann Equation) Small Potential Approximation: q Φ kT << 1 sinh q Φ q Φ	https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf
∇ Φ − Φ 2 = 0 L d 1 ∂ ⎛ 2 ∂Φ ⎞ r r2 ∂r ⎜ ∂r ⎟ ⎠ ⎝ = 1 ∂2 r ∂r 2 (r Φ) E. Ohmic Conduction ⇒ d2 ) ( dr2 r Φ − r Φ L2 d = 0 0 r Φ = A e 1 −r / Ld + A e 2 +r / L d Φ ( )r = Q 4 π ε r −r / L d e J = ρ µ E + + + − D + ∇ρ + J = −ρ µ E − − − − D − ∇ρ − If charge density gradients small, then ∇ρ ± negligible ⇒ ρ + = −ρ − = ρ o J = J+ + J− = (ρ µ − ρ µ ) E = ρ (µ + µ ) E = σE − − o + + − + J = σ E (Ohm’s Law) σ = ohmic conductivity 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 13 of 27 F. pn Junction Diode 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 14 of 27 ∆Φ = Φ − Φ = n p k T N N ln 2 ni q A D Φ (x = 0 )= Φ p + 2 q NA xp 2ε = Φ n − 2 q ND xn 2ε ∆Φ = Φ − Φ = n p 2 q ND xn 2ε + 2 q NA xp 2ε =	https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf
− Φ = n p 2 q ND xn 2ε + 2 q NA xp 2ε = q ND xn 2ε (xn + xp ) = q N D xn 2ε 2 ⎛ ⎜1 + ⎝ ND ⎞ ⎟ NA ⎠ 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 15 of 27 VII. Relationship Between Resistance and Capacitance In Uniform Media Described by ε and σ . C = uq v = S (cid:118) ∫ D da i ∫ E ds i L = ε ∫ (cid:118) E da i S ∫ E ds i L i ∫ E ds v L R = = i i(cid:118) ∫ J da S = L σ ∫ E ds i i(cid:118) ∫ E da S RC = L σ ∫ E ds i i(cid:118) ∫ E da S ε i (cid:118) ∫ E da = L (cid:118) ∫ E ds i ε σ L Check: Parallel Plate Electrodes: R = l Aσ , ε A C = l ⇒ ε RC = σ 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 16 of 27 Coaxial R = ln b a 2 π σ l , C = 2 π ε l b ln a ε ⇒ RC = σ Concentric Spherical 1 R 1 − 1 R 2 4 π σ R = , C = 4 π ε 1 1 − R 1 R 2	https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf
R 2 4 π σ R = , C = 4 π ε 1 1 − R 1 R 2 ε ⇒ RC = σ 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 17 of 27 VIII. Change Relaxation in Uniform Conductors ∇ i Ju + ∂ ρ ∂ t u = 0 ρ ∇ i E = u ε J = u σ E σ ∇ i E + ∂ ρ ∂ t u = 0 ⇒ ∂ ρ u ∂ t + σ ε ρ = 0 u ρ u ε ∂ ρu ∂ t + τ = e σε = dielectric relaxation time ρu = 0 ⇒ ρ = ρ (r, t = 0 ) e − t τ 0 u τ e e IX. Demonstration 7.7.1 – Relaxation of Charge on Particle in Ohmic Conductor Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 18 of 27 (cid:118) ∫ J da = i σ (cid:118) ∫ E da = i S S σ qu ε = −dq dt dq q dt τe + = 0 ⇒ q = q t = 0 e − τ t e ) ( (τe = σ) ε Partially Uniformly Charged Sphere Courtesy of Krieger Publishing. Used with permission. 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 19 of 27 ρu (t = 0) = ρ0 r	https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf
. Markus Zahn Lecture 7 Page 19 of 27 ρu (t = 0) = ρ0 r R 1< 0 r > R 1 Q T = 4 3 π R 1 3 ρ0 ( ) ρu t = ρ e t 0 − τ e r < R 1 (τ = ε σ) e 0 r > R 1 − τ ρ0 r e t e 3 ε = Q r e t e − τ 4 π ε R 3 1 0 < r < R 1 r ( E r, t ) = Q e t e − τ 4 π ε r 2 Q 4 π ε0 r 2 R < 1 r R < 2 r > R 2 σ su ) ( r = R = ε0 E r = R 2 − ε E r = R 2 − r ( 2 ) + ) r ( = Q 4 R 2π 2 (1 e t e − − τ ) X. Self-Excited Water Dynamos A. DC High Voltage Generation (Self-Excited) 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 20 of 27 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission. Courtesy of Herbert Woodson and James Melcher. Used with permission. Woodson, Herbert H., and James R. Melcher. Electromechanical Dynamics, Part 2: Fields, Forces, and Motion. Malabar, FL: Kreiger Publishing Company, 1968. ISBN: 9780894644597. 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 21 of 27 − n C v = C 1 i −n C v = C 2 i dv2 dt dv1 dt v = V (cid:108) est 1	https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf
�� ⎢ ⎥ ⎥ ⎢ ⎥ ⎥ ⎢V(cid:108) ⎥ ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎥ V(cid:108) n C i 3 ⎥ ⎦ ⎢ ⎦ ⎣ 2 = 0 ⎡n Ci ⎢ ⎢ ⎢0 ⎢ ⎢ ⎢ ⎣Cs Cs n C i 0 det = 0 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission. 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 22 of 27 (n Ci )3 + Cs )3 ( = 0 ⇒ s = ⎜ ⎛ n Ci ⎞ ( ⎟ ⎝ C ⎠ −1 )1 3 s = −n C C (exponentially decaying solution) 1 i (−1)1 3 = −1, 1 ± 3j 2 s = 2, 3 n Ci ⎡1 ± 3 j ⎤ (blows up exponentially because sreal >0 ; but also 2 C ⎣ ⎦ oscillates at frequency simag ≠ 0) XI. Conservation of Charge Boundary Condition ∇ i Ju + ∂ρ ∂t u = 0 J da + (cid:118)∫ u i S d dt ∫ ρu dV = 0 V ⎣ a − Jb ⎤ i ⎡ n J ⎦ + d dt σsu = 0 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 23 of 27	https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf
, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 23 of 27 XII. Maxwell’s Capacitor A. General Equations _ ( ) x E t i a 0 < x < a _ ( ) x E t i b − b x < 0 < E = a ∫ x E d = v t = E t b + E t a a x b ( ) ( ) ( ) −b ⎡ i ⎣ a n J − Jb ⎤ + ⎦ dσsu = 0 ⇒ σ E ( ) − σ E ( ) + a t dt b t b a d dt ⎣ a a ( ) ⎡ε E t − ε E ( ) = 0 b b t ⎦⎤ ( ) v t ( ) E t = b b − E ( ) a t b a σ a E t a ⎡ ( ) v t ( ) − σ ⎢ ⎢ b ⎣ b − ⎤ ( ) E t a ⎥ + ⎥ dt ⎦ d ⎡ ( ) ε E t ⎢ ⎢ ⎣ a a a − ε ( ) v t ⎛ ⎜ b ⎜ b ⎝ − ⎞⎤ E t a ⎟⎥ 0 ⎟ ⎥ ⎠⎦ ( ) = a ⎛ ε ⎜ a ⎝ + b ε a dE ⎞ ⎟ b ⎠ dt a ⎛ + σ + ⎜ a ⎝ σ a ⎞ b E t ⎟ a b	https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf
+ σ + ⎜ a ⎝ σ a ⎞ b E t ⎟ a b ⎠ ( ) = ( ) ε b v t σ b dv + b dt b 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 24 of 27 B. Step Voltage: v t ( ) = V u t ( ) Then dv dt = V δ t (an impulse) ( ) At t=0 ⎛ ⎜εa + ⎝ εb εba dE a = dv ⎞ ⎟ b dt b ⎠ dt εb Vδ ( ) t = b Integrate from t=0- to t=0+ t 0 = ε a dE + ⎛ ⎜εa + b ⎞ ⎟ b ⎠ dt ∫ t 0= − ⎝ ⎛ ⎜εa + a dt = ⎝ εba ⎞ b ⎠ ⎟ Ea t 0+ = t 0= − = 0 + ∫ t 0 = − ε b b ( ) Vδ t dt = ε b b V E t( = 0 − ) = 0 a ⎛ ε ⎜ a ⎝ + εba ⎞ b ⎠ ) = E t = 0 ( ⎟ a + εb b For t > 0, dv dt = 0 ) = ( V ⇒ E t = 0 a + b ε V εb b + εb a ⎛ ε + ⎜ a ⎝ b ε a dE ⎞ ⎟ b ⎠ dt a ⎛	https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf
σ + ω ε + j Ea ⎢ a ⎣ a a (σ + ω b j ε )⎤ ⎥ ⎦ b b = V(cid:108) b ⎣ b ⎡σ + ω ε ⎤ = 0 j b ⎦ E(cid:3) j ω ε + σ a b b = E(cid:3) j ω ε + σ b a a = V(cid:108) ⎡b (σ + ω ⎣ j ε ) + a (σ + ω ε )⎤ b ⎦ j a a b σ(cid:3) su = ε E(cid:3) − ε E(cid:3) b a b a = (ε σ − ε σ ) V(cid:108) b (σ + ω ε ) + a ⎡ j ⎣ a b a b b a a (σ + ω j ε )⎤ b ⎦ 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 26 of 27 D. Equivalent Circuit (Electrode Area A) (cid:3)I = σ + j ω ε E A = σ + j ω ε E A b ) (cid:3) a ) (cid:3) ( a ( b b a = R a R C j a a V(cid:108) + R b ω + 1 R C j ω + 1 b b R = a a σa A , R = b b σb A ε A C = a a a ε A , C = b b b 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7	https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf
, C = b b b 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 7 Page 27 of 27	https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf
18.997 Topics in Combinatorial Optimization March 4, 2004 Lecture 8 Lecturer: Michel X. Goemans Scribe: Constantine Caramanis This lecture covers the proof of the Bessy-Thomass´e Theorem, formerly known as the Gallai Conjecture. Also, we discuss the cyclic stable set polytope, and show that it is totally dual integral (TDI) (see lecture 5 for more on TDI systems of inequalities). 1 Recap and Deﬁnitions In this section we provide a brief recap of some deﬁnitions we saw in the previous lecture. Also we answer a question that remained unanswered in the previous lecture regarding the polynomiality of ﬁnding a valid ordering given any strongly connected directed graph. For a strongly connected digraph D = (V, A), with \|V \| = n, we make the following deﬁnitions. 1. Given an enumeration of the vertices, {v1, . . . , vn}, an arc (vi, vj ) ∈ A is called backward if i > j and forward if i < j. 2. An ordering O, is an equivalence class of enumerations of a graph. The equivalence class is deﬁned by the equivalence relations (a) v1, v2, . . . , vn ∼ v2, v3, . . . , vn, v1, (b) v1, v2, . . . , vn ∼ v2, v1, v3, . . . , vn, if there is no arc between v1 and v2, i.e., (v1, v2), (v2, v1) /∈ A. 3. Given an ordering O, the index with respect to O of a directed cycle C, denoted iO(C), is the number of backward arcs in C. Recall from the last lecture that the	https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf
cycle C, denoted iO(C), is the number of backward arcs in C. Recall from the last lecture that the index is well deﬁned, since the index is invariant under the equivalence operations deﬁned above. 4. We say that an ordering O is valid if for any arc (u, v) ∈ A, there exists a cycle C containing that arc, with index 1: iO (C) = 1. We showed in the last lecture that there always exists a valid ordering. 5. A cyclic stable set S with respect to a valid ordering O, is such that S is a stable set on the underlying undirected graph, and also there exists some enumeration {v1, . . . , vn} of the ordering such that S = {v1, . . . , vk }, where k = \|S\|. Last time we proved that any strongly connected digraph has a valid ordering. In fact, given any such graph, a valid ordering can be found in time polynomial in the size of the graph. Recall that the proof of the existence theorem showed that the minimizer of min O � iO(C), directed cycles C must be a valid ordering. Given any ordering O, we showed in the proof that in a polynomial number of steps (essentially, by repeated “local swaps”), if O is not valid, we can obtain a new ordering O1, reducing the number of arcs for which there are no cycles of index 1 containing them. Therefore we can ﬁnd a valid ordering in polynomial time. 8-1 2 The Bessy-Thomass´e Theorem Recall the statement of the theorem. Theorem 1 Given a strongly connected digraph D = (V, A), and a valid ordering O, if αO denotes the size of the largest cardinality	https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf
, A), and a valid ordering O, if αO denotes the size of the largest cardinality cyclic stable set, then � αO = min iO(Ci), where the cycles {C1, . . . , Cp} cover the vertex set V . {C1,...,Cp} The inequality αO ≤ min � iO(Ci), {C1,...,Cp} is straightforward (as each vertex of a cycle stable set must be contained in (at least) one directed cycle and the corresponding entering arc must be backward), so we consider only the proof of the reverse inequality. Before we prove this theorem, we make some remarks. It is important to note that the cyclic stability number, αO , depends on the ordering O chosen. To illustrate this, recall our digraph on ﬁve vertices from last lecture. In Figure 2, we exhibit two diﬀerent orderings where the cyclic stability number is diﬀerent. � � � � � � � � � � � � � � � � � � � � � � � � � � � � Figure 1: In ﬁgure (a) above, the cyclic stability number equals 2, where as in (b), the cyclic stability number equals 1. Computing the stability number of a general graph is known to be N P -hard. One of the corollar- ies of the Bessy-Thomass´e Theorem is that the cyclic stability number can be computed eﬃciently. iO(Ci) in the right hand This follows because we can can compute the quantity min side of the	https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf
the right hand This follows because we can can compute the quantity min side of the theorem above, eﬃciently. We can do this by formulating a network ﬂow problem that computes the minimization. To do this, ﬁx an enumeration of the ordering. Attach a cost of 0 to every forward arc in the digraph under the given enumeration, and a cost of 1 to every backward arc. Next, split each vertex v into a pair {vout, v in) with ﬂow capacity bounded from below by 1. Then for every arc (u, v) in the original graph, draw an arc (uout, vin) in the network ﬂow graph. Finding a minimum cost ﬂow in this network can be done eﬃciently, and it amounts to ﬁnding a set of cycles {C1, . . . , Cp} that cover V , and minimize } with a directed edge (vout, v {C1,...,Cp} � � iO (Ci). in {C1,...,Cp} A key step in the proof of the Bessy-Thomass´e Theorem is a lemma that provides a suﬃcient condition for a subset S of vertices to be a cyclic stable set. 8-2 Lemma 2 Given a valid ordering O, ﬁx an enumeration, {v1, . . . , vn}. Let S ⊆ V be a subset of the vertices. If there are no forward paths between any two vertices of S, then S is a cyclic stable set. Suppose, to the contrary, that S has no forward arcs, but S	https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf
is a cyclic stable set. Suppose, to the contrary, that S has no forward arcs, but S is not a cyclic stable. Let vi Proof: be the ﬁrst element of the enumeration in S. If we rotate the enumeration so that vi becomes v1, no forward paths are either created or destroyed in S, so we may assume, without loss of generality, that v1 ∈ S. If S is a cyclic stable set with respect to O, then there exists some enumeration of O for which the elements of S are the ﬁrst k = \|S\| elements of the enumeration. Equivalently, there exists a sequence of local steps, or swaps we can make according to the equivalence relations deﬁning an ordering, to move from the current enumeration to one of the correct form. If S is not a cyclic stable set, as we assume, then this is not possible. Consider the enumeration which brings S “as close as possible” to having all its elements at the beginning of the enumeration, as illustrated in Figure 2. By this we mean that as many elements of S as possible are listed ﬁrst in the enumeration, and furthermore, the ﬁrst element of S not part of the initial string of elements of S (which we call S<) is as close to S< as possible. We denote by S< the elements of S that are at the beginning of � � � � � � � � � � � � � � � � � � � � Figure 2: The ﬁgure exhibits the enumeration with respect to which as many elements of S as possible are the ﬁrst	https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf
enumeration with respect to which as many elements of S as possible are the ﬁrst elements of the enumeration. Since S is assumed not to be a cyclic stable set, there must be some element w sandwiched by elements of S. the enumeration, by S> the remaining elements of S, and by W the elements after the last element of S< and before the ﬁrst element of S>, as illustrated in Figure 2. Since there are no forward paths joining any two elements of S, for any w ∈ W there cannot be a forward path from S< to w, and a forward path from w to an element of S>. Consider the ﬁrst w ∈ W where there is no forward path from S< to w (if there is such a vertex). Because w is assumed to be the ﬁrst such vertex, there can be no forward path from any vertex v coming before w in the enumeration. If there were such a vertex v, then if there were a forward path from S< to v, we would also have a forward path from S< to w. If there were no forward path from S< to v, it would contradict our assumption that w is the ﬁrst vertex in W that has no forward path from S<. In particular, then, there are no arcs from any vertex before w in the enumeration, to w. However, there also can be no arc from w to any vertex before	https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf
, to w. However, there also can be no arc from w to any vertex before it in the enumeration. This follows because O was assumed to be a valid ordering. If there were such an arc, say (w, v) for v earlier in the enumeration, because we assume there are no forward arcs from any vertex coming before w, to w, and that w is the ﬁrst such vertex, then any cycle C containing the arc (w, v) must have iO(C) ≥ 2, a contradiction to the validity of the ordering O. Therefore there are no arcs between w and any vertex previous to w in the enumeration. But then using the equivalence relations, we can swap w with each element before it, including then each element of S<. But this contradicts our assumption that the ﬁrst element of 8-3 S \ S< was as close as possible to S<. Therefore there are no elements in W that have no forward paths from S. In particular, this implies that there are no forward paths from any w ∈ W to S>. Then, let vj ∈ S be the ﬁrst vertex in S>. By assumption, unless W is empty, vj−1 ∈ W , and there is no forward path from vj−1 to S>, and in particular, (vj−1, vj ) /∈ A. But then, since O is a valid ordering, (vj , vj−1) /∈ A. In this case, we can swap the two vertices, contradicting our assumption that our enumeration put S “as close as possible” to having its elements at the beginning of the � enumeration. Therefore W must be empty, and S is indeed a cyclic stable set. We now move to the proof of the B	https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf
enumeration. Therefore W must be empty, and S is indeed a cyclic stable set. We now move to the proof of the Bessy-Thomass´e Theorem. Proof: The Main Idea: We want to show that the size of the maximum cyclic stable set equals the minimum total index of a family of cycles covering V . Essentially the proof relies on mapping our digraph D to a poset T . At this point, we appeal to Dilworth’s Theorem (lecture 6). Recall that the strong version of Dilworth’s Theorem tells us that the size of the largest antichain in the poset equals the minimum number of chains needed to partition the elements of the poset. We show that our maximum size cyclic stable set S in D, corresponds naturally to an antichain in the poset T . Thus the size of the largest antichain in T is at least the size of S, i.e., αO. Then we use Lemma 2 to show that any antichain in T corresponds to a cyclic stable set in D. Thus we have that the size of the largest antichain in T is exactly αO . Dilworth’s Theorem now links the number of chains partitioning T to αO . The ﬁnal part of the proof recovers a covering family of cycles from the chains in T . The Proof: For the given ordering O, let S = {v1, . . . , vk } denote the maximum size cyclic sta- ble set, with corresponding enumeration of V = {v1, . . . , vk , . . . , vn}. We note that since S is, in particular, a stable set, we can permute its elements as we wish within the given ordering. (cid:1) We form an acyclic digraph D(cid:1) = (V (	https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf
the given ordering. (cid:1) We form an acyclic digraph D(cid:1) = (V (cid:1), A(cid:1)) from D as follows. Let V (cid:1) = {v1, . . . , vn, v1, . . . , vk (cid:1) } (we duplicate the elements of S) so that \|V (cid:1)\| = n + k. Next, if (v, w) ∈ A is a forward arc, then (v, w) ∈ A(cid:1). If (w, vi ) ∈ A is a backward arc into a vertex of S (i.e., if vi ∈ S) then (w, vi ) ∈ A(cid:1) . Note that by our choice of enumeration, any arcs into vi, i ≤ k, must be backward. Therefore the digraph D(cid:1) is acyclic. It is illustrated in Figure 2. (cid:1) � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � Figure 3: This ﬁgure illustrates the directed acyclic graph D� we obtain from splitting the vertices in S and drawing arcs as explained above. In order to use Dilworth’s Theorem, we need to have a poset T . We obtain a poset T from the acyclic digraph D(cid:1) by considering the transitive closure of D(cid:1) . Since the sets {v1, . . . , vk } and 8-4 (cid:1) (cid:1) {v	https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf
, . . . , vk } and 8-4 (cid:1) (cid:1) {v1, . . . , vk } have no incoming and outgoing arcs, respectively, they are both antichains in T . This is also evident from Figure 2. We show that they are in fact maximum size antichains. Consider any antichain I. As the ordering is valid, for any vertex, there exists a directed cycle of index 1 going through it. This translates into a chain in the poset going from vi to vi for any 1 ≤ i ≤ k. This means that an antichain I cannot contain both vi and vi. Let ID be the elements of the original digraph D corresponding to I. (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) / (cid:1) l (cid:1) k By renumbering the elements of S (recall that we can permute the elements of S within the given ordering) we can assume that v1, . . . , v ∈ I, and vl+1, . . . , v ∈ I. Now rotate the enumeration to obtain {v˜1, . . . , v˜n} so that v˜1 = vl+1. Since I is an antichain, and since the digraph vertices (cid:1) {v1, . . . , vl} corresponding to the poset elements {v1, . . . , vl } at the “top” of the poset T have been rotated to be the last elements of the enumeration, there are no forward paths between any two elements of ID . Therefore, by Lemma 2, ID	https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf
are no forward paths between any two elements of ID . Therefore, by Lemma 2, ID is a cyclic stable set. Therefore ID , and consequently I, can have size at most equal to the size of S, that is, αO . We have thus shown that the size of the largest antichain in T is equal to the cyclic stability number αO of D. Now consider the minimal partitioning set of chains in the poset T , call these P1, . . . , Pk (where k = \|S\| = αO ). Each chain Pi is a chain from vi to vσ(i), for some permutation σ of {1, . . . , k}. By a slight abuse of notation, we also use Pi to refer to the directed path in D from vi to vσ(i) (or cycle if σ(i) = i). We note that by construction of T , there is exactly one backward arc in each path Pi, namely, the last arc to vσ(i). These paths cover the vertex set V . Now, the cycles in the permutation σ correspond to cycles in D. For example, if (12) is a cycle in σ, i.e., if σ(1) = 2 and σ(2) = 1, then joining the paths P1 and P2 we have a cycle from v1 to v1. We note that these cycles may in fact intersect. Since the cycles merely need to cover the vertex set V , distinct cycles can intersect. We need to take care that the same cycle does not intersect itself. If σ happens to be the identity permutation, σ(i) = i, then each path is a cycle and cannot intersect itself, and hence the proof is complete. If this is not the case, then a cycle in D obtained by	https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf
proof is complete. If this is not the case, then a cycle in D obtained by joining together the paths Pi that correspond to a cycle of σ may in fact intersect itself. Suppose that i and j are in the same cycle of σ and the paths Pi and Pj intersect, in say v. We can then replace the paths Pi and Pj by two other paths P (cid:1) and Pj (cid:1) (obtained by switching from one to the other at v) which together cover the same vertices and which corresponds to a new permutation σ(cid:1) with σ(cid:1)(i) = σ(j) and σ(cid:1)(j) = σ(i). Now the number of cycles in the permutation has increased by one, and we can repeat this process until no cycle in D (corresponding to each cycle of teh permutation σ) intersects itself. i Since the cycle splitting procedure does not change the total index of the cycles, we know that the total index equals the minimal number of chains required to partition T . But by above, this is exactly the size of the maximum cyclic stable set, and therefore αO = min � iO(Ci), {C1,...,Cp} which is what we wanted to prove. � 3 Cyclic Stable Set Polytope In this section, we follow some recent (unpublished) work of A. Seb¨o, and deﬁne the cyclic stable set polytope of a strongly connected graph D, with a given valid ordering O. Deﬁne the polytope P as follows. � (cid:2) � � x(C) ≤ iO(C), ∀ directed cycles C P = x � xv ≥ 0, ∀v	https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf
x(C) ≤ iO(C), ∀ directed cycles C P = x � xv ≥ 0, ∀v ∈ V � . We show in this section that the polytope P is totally dual integral (TDI) (see lecture 5 for more on TDI system of inequalities). Given a cyclic stable set S (cyclic stable with respect to the given ordering), let xS denote its incidence vector, i.e., xv = 1 if v ∈ S, and 0 otherwise. Then in fact xS ∈ P. Indeed, consider any S 8-5 directed cycle C. Since S is cyclic stable, C always enters S via a backward arc, and therefore the number of backward arcs of C is at least the cardinality of its intersection with S: (# backward arcs in C) = iO(C) ≥ \|C ∩ S\|, or, equivalently, xS (C) ≤ iO(C). Since we have shown that the incidence vector of every cyclic stable set belongs to P, we have: αO ≤ max : s.t. : � xv v∈V x(C) ≤ iO(C), ∀C xv ≥ 0, ∀v ∈ V By linear programming duality, and then by observing that the optimum value of a minimization can only increase if we add constraints, we have � αO ≤ max : xv v∈V s.t. : x(C) ≤ iO(C), ∀C xv ≥ 0, ∀v ∈ V � = min : � C iO(C)yC yC ≥ 1, ∀v ∈ V s.t. : C : v∈C yC ≥ 0, ∀C � ≤ min : � C s.t. : iO(C)yC C : v∈C yC	https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf
� ≤ min : � C s.t. : iO(C)yC C : v∈C yC ≥ 0, ∀C yC ∈ {0, 1}. yC ≥ 1, ∀v ∈ V But this last quantity is exactly the minimum total index of a cycle cover of V , and thus by the Bessy-Thomass´e Theorem, the ﬁnal quantity equals αO . Therefore equality must hold throughout. Recall that in order to prove that the description of P is TDI, we must show that for all integral objective functions w (wv ∈ Z), the dual linear program � min : � C s.t. : iO(C)yC C : v∈C yC ≥ 0, ∀C yC ≥ 1, ∀v ∈ V has an integral solution whenever its value is ﬁnite. We note that we have just proved this statement for the special case wv = 1. We note also that if we have wv ≤ 0, we can replace this wv by 0 without aﬀecting the feasible region of the dual linear program. Therefore, we can assume that we have wv ∈ Z+. We now construct a strongly connected digraph D(cid:1) = (V (cid:1), A(cid:1)), with valid ordering O(cid:1) as follows. Let V (cid:1) consist of wv copies of each xv , {xv,1, . . . , xv,wv } (recall that wv is a positive integer). If (v, u) ∈ A, then (xv,i, xu,j ) ∈ A(cid:1) for every i ≤ wv and j ≤ wu. From our reasoning above, we know that the linear program associated to the digraph D(cid	https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf
≤ wu. From our reasoning above, we know that the linear program associated to the digraph D(cid:1) (now we have wv = 1 for every v ∈ V (cid:1)) produces an integral solution that corresponds to a maximum size cyclic stable set in D(cid:1). Note that if xv,i is in the stable set S(cid:1) for D(cid:1), then we can also take xv,j to be in S(cid:1) for any j ≤ wv . Therefore any maximum size cyclic stable set S(cid:1) in D(cid:1) naturally corresponds to a cyclic stable set S in D. Moreover, \|S(cid:1)\| = w xS . Conversely, if S is a cyclic stable set in D, then the set S(cid:1) of all copies of the vertices in S, is a cyclic stable set in D(cid:1), with \|S(cid:1)\| = w xS . Therefore given any vector w with wv ∈ Z+, the linear program with objective function w x has an integral optimal solution. Therefore P is totally dual integral, as we wished to show. (cid:1) (cid:1) (cid:1) 8-6	https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.685 Electric Machines Class Notes 1: Electromagnetic Forces c 2003 James L. Kirtley Jr. (cid:13) 1 Introduction Bearings Shaft End Windings Stator Stator Conductors Rotor Air Gap Rotor Conductors Figure 1: Form of Electric Machine This section of notes discusses some of the fundamental processes involved in electric machinery. In the section on energy conversion processes we examine the two major ways of estimating elec- tromagnetic forces: those involving thermodynamic arguments (conservation of energy) and ﬁeld methods (Maxwell’s Stress Tensor). But ﬁrst it is appropriate to introduce the topic by describing a notional rotating electric machine. Electric machinery comes in many diﬀerent types and a strikingly broad range of sizes, from those little machines that cause cell ’phones and pagers to vibrate (yes, those are rotating electric machines) to turbine generators with ratings upwards of a Gigawatt. Most of the machines with which we are familiar are rotating, but linear electric motors are widely used, from shuttle drives in weaving machines to equipment handling and amusement park rides. Currently under development are large linear induction machines to be used to launch aircraft. It is our purpose in this subject to develop an analytical basis for understanding how all of these diﬀerent machines work. We start, however, with a picture of perhaps the most common of electric machines. 2 Electric Machine Description: Figure 1 is a cartoon drawing of a conventional induction motor. This is a very common type of electric machine and will serve as a reference point. Most other electric machines operate in 1 a fashion which is the same as the induction machine or which diﬀer in ways which are easy to reference to the induction machine. Most (but not all!) machines we will be studying have essentially this morphology. The rotor of the machine is mounted on a shaft which is supported on some sort of bearing(s). Usually, but not always, the rotor is inside. I have drawn a rotor which is round, but this does not need to be the case. I have also indicated rotor conductors, but sometimes the rotor has permanent magnets either fastened to it or inside, and sometimes (as in Variable Reluctance Machines)	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
indicated rotor conductors, but sometimes the rotor has permanent magnets either fastened to it or inside, and sometimes (as in Variable Reluctance Machines) it is just an oddly shaped piece of steel. The stator is, in this drawing, on the outside and has windings. With most of the machines we will be dealing with, the stator winding is the armature, or electrical power input element. (In DC and Universal motors this is reversed, with the armature contained on the rotor: we will deal with these later). In most electrical machines the rotor and the stator are made of highly magnetically permeable materials: steel or magnetic iron. In many common machines such as induction motors the rotor and stator are both made up of thin sheets of silicon steel. Punched into those sheets are slots which contain the rotor and stator conductors. Figure 2 is a picture of part of an induction machine distorted so that the air-gap is straightened out (as if the machine had inﬁnite radius). This is actually a convenient way of drawing the machine and, we will ﬁnd, leads to useful methods of analysis. Stator Core Stator Conductors In Slots Air Gap Rotor Conductors In Slots Figure 2: Windings in Slots What is important to note for now is that the machine has an air gap g which is relatively small (that is, the gap dimension is much less than the machine radius r). The air-gap also has a physical length l. The electric machine works by producing a shear stress in the air-gap (with of course side eﬀects such as production of “back voltage”). It is possible to deﬁne the average air- gap shear stress, which we will refer to as τ . Total developed torque is force over the surface area times moment (which is rotor radius): T = 2πr2ℓ < τ > Power transferred by this device is just torque times speed, which is the same as force times 2 surface velocity, since surface velocity is u = rΩ: Pm = ΩT = 2πrℓ < τ > u If we note that active rotor volume is , the ratio of torque to volume is just: T Vr = 2 < τ > Now, determining what can be done in a volume of machine involves two things. First,	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
is just: T Vr = 2 < τ > Now, determining what can be done in a volume of machine involves two things. First, it is clear that the volume we have calculated here is not the whole machine volume, since it does not include the stator. The actual estimate of total machine volume from the rotor volume is actually quite complex and detailed and we will leave that one for later. Second, we need to estimate the value of the useful average shear stress. Suppose both the radial ﬂux density Br and the stator surface current density Kz are sinusoidal ﬂux waves of the form: Br = √2B0 cos (pθ Kz = √2K0 cos (pθ ωt) ωt) − − Note that this assumes these two quantities are exactly in phase, or oriented to ideally produce torque, so we are going to get an “optimistic” bound here. Then the average value of surface traction is: < τ >= BrKzdθ = B0K0 2π 1 2π 0 Z The magnetic ﬂux density that can be developed is limited by the characteristics of the magnetic materials (iron) used. Current densities are a function of technology and are typically limited by how much eﬀort can be put into cooling and the temperature limits of insulating materials. In practice, the range of shear stress encountered in electric machinery technology is not terribly broad: ranging from a few kPa in smaller machines to about 100 kPa in very large, well cooled machines. It is usually said that electric machines are torque producing devices, meaning tht they are deﬁned by this shear stress mechanism and by physical dimensions. Since power is torque times rotational speed, high power density machines necessarily will have high shaft speeds. Of course there are limits on rotational speed as well, arising from centrifugal forces which limit tip velocity. Our ﬁrst step in understanding how electric machinery works is to understand the mechanisms which produce forces of electromagnetic origin. 3 Energy Conversion Process: In a motor the energy conversion process can be thought of in simple terms. In “steady state”, electric power input to the machine is just the sum of electric power inputs to the diﬀerent phase terminals: Mechanical power is torque times speed: Pe = viii	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
the sum of electric power inputs to the diﬀerent phase terminals: Mechanical power is torque times speed: Pe = viii i X Pm = T Ω 3 Electric Power In Electro- Mechanical Converter Mechanical Power Out Losses: Heat, Noise, Windage,... Figure 3: Energy Conversion Process And the sum of the losses is the diﬀerence: Pd = Pe Pm − It will sometimes be convenient to employ the fact that, in most machines, dissipation is small enough to approximate mechanical power with electrical power. In fact, there are many situations in which the loss mechanism is known well enough that it can be idealized away. The “thermodynamic” arguments for force density take advantage of this and employ a “conservative” or lossless energy conversion system. 3.1 Energy Approach to Electromagnetic Forces: + v - Magnetic Field System f x Figure 4: Conservative Magnetic Field System To start, consider some electromechanical system which has two sets of “terminals”, electrical and mechanical, as shown in Figure 4. If the system stores energy in magnetic ﬁelds, the energy stored depends on the state of the system, deﬁned by (in this case) two of the identiﬁable variables: ﬂux (λ), current (i) and mechanical position (x). In fact, with only a little reﬂection, you should be able to convince yourself that this state is a single-valued function of two variables and that the energy stored is independent of how the system was brought to this state. Now, all electromechanical converters have loss mechanisms and so are not themselves conser- vative. However, the magnetic ﬁeld system that produces force is, in principle, conservative in the 4 sense that its state and stored energy can be described by only two variables. The “history” of the system is not important. It is possible to chose the variables in such a way that electrical power into this conservative system is: P e = vi = i dλ dt Similarly, mechanical power out of the system is: P m = f e dx dt The diﬀerence between these two is the rate of change of energy stored in the system	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
is: P m = f e dx dt The diﬀerence between these two is the rate of change of energy stored in the system: dWm = P e dt − P m It is then possible to compute the change in energy required to take the system from one state to another by: Wm(a) Wm(b) = a idλ f edx − where the two states of the system are described by a = (λa, xa) and b = (λb, xb) b Z − If the energy stored in the system is described by two state variables, λ and x, the total diﬀerential of stored energy is: and it is also: dWm = ∂Wm ∂λ dλ + ∂Wm ∂x dx dWm = idλ f edx − So that we can make a direct equivalence between the derivatives and: f e = ∂ m W ∂x − In the case of rotary, as opposed to linear, motion, torque T e takes the place of force f e and angular displacement θ takes the place of linear displacement x. Note that the product of torque and angle has the same units as the product of force and distance (both have units of work, which in the International System of units is Newton-meters or Joules. In many cases we might consider a system which is electricaly linear, in which case inductance is a function only of the mechanical position x. In this case, assuming that the energy integral is carried out from λ = 0 (so that the part of the integral carried out over x is zero), λ(x) = L(x)i Wm = λ 1 0 L(x) Z λdλ = 1 λ2 2 L(x) This makes f e = 1 2 − λ2 ∂ 1 ∂x L(x) 5 Note that this is numerically equivalent to f e = 1 i2 ∂ 2 ∂x − L(x) This is true only in the case of a linear system. Note that substituting L(x)i = λ too early in the derivation produces erroneous results: in the case of a linear system it produces a sign error, but in	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
L(x)i = λ too early in the derivation produces erroneous results: in the case of a linear system it produces a sign error, but in the case of a nonlinear system it is just wrong. 3.1.1 Example: simple solenoid Consider the magnetic actuator shown in cartoon form in Figure 5. The actuator consists of a circular rod of ferromagnetic material (very highly permeable) that can move axially (the x- direction) inside of a stationary piece, also made of highly permeable material. A coil of N turns carries a current I. The rod has a radius R and spacing from the ﬂat end of the stator is the variable dimension x. At the other end there is a radial clearance between the rod and the stator g. Assume g R. If the axial length of the radial gaps is ℓ = R/2, the area of the radial gaps is the same as the area of the gap between the rod and the stator at the variable gap. ≪ R/2 x µ R g CL µ N turns Figure 5: Solenoid Actuator The permeances of the variable width gap is: µ 0πR2 x 1 = P and the permeance of the radial clearance gap is, if the gap dimension is small compared with the radius: The inductance of the coil system is: 2 = P 2µ0πRℓ g = 0 R2 µ π g L = N 2 R1 + R2 = N 2 P P = P2 + 1 2 P2 2 µ 2 0πR N x + g Magnetic energy is: 6 Wm = 0 Z λ 0 idλ = 1 λ2 2 L(x) λ2 x + g = 0 2 µ πR2N 2 0 And then, of course, force of electric origin is: Here that is easy to carry out: So that the force is: f e = m ∂W ∂x − = λ2 0 d − 2 dx L(x) 1 d 1 dx L = 1 2 0πR N µ 2 f e (x)	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
dx L(x) 1 d 1 dx L = 1 2 0πR N µ 2 f e (x) = λ2 0 2 µ 1 0πR N 2 2 − Given that the system is to be excited by a current, we may at this point substitute for ﬂux: λ = L(x)i = µ0πR2N i x + g and then total force may be seen to be: f e = 0πR2N 2 i2 µ (x + g)2 2 − The force is ‘negative’ in the sense that it tends to reduce x, or to close the gap. 3.1.2 Multiply Excited Systems There may be (and in most electric machine applications there will be) more than one source of electrical excitation (more than one coil). In such systems we may write the conservation of energy expression as: dWm = ikdλk k X f edx − which simply suggests that electrical input to the magnetic ﬁeld energy storage is the sum (in this case over the index k) of inputs from each of the coils. To ﬁnd the total energy stored in the system it is necessary to integrate over all of the coils (which may and in general will have mutual inductance). Of course, if the system is conservative, Wm(λ1, λ2, . . . , x) is uniquely speciﬁed and so the actual path taken in carrying out this integral will not aﬀect the value of the resulting energy. Wm = dλ i · Z 7 3.1.3 Coenergy We often will describe systems in terms of inductance rather than its reciprocal, so that current, rather than ﬂux, appears to be the relevant variable. It is convenient to derive a new energy variable, which we will call co-energy, by: W ′ m = λiii − Wm i X and in this case it is quite easy to show that the energy diﬀerential is (for a single mechanical variable) simply: so that force produced is: dW ′ m = λkdik + f dx e k X fe = ∂W ′	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
that force produced is: dW ′ m = λkdik + f dx e k X fe = ∂W ′ m ∂x 3.2 Example: Synchronous Machine Stator Gap Rotor C A’ B θ B’ A C’ F’ µ F F F’ C’ A B’ B µ C A’ Figure 6: Cartoon of Synchronous Machine Consider a simple electric machine as pictured in Figure 6 in which there is a single winding on a rotor (call it the ﬁeld winding and a polyphase armature with three identical coils spaced at uniform locations about the periphery. We can describe the ﬂux linkages as: λa = Laia + Labib + Labic + M cos(pθ)if λb = Labia + Laib + Labic + M cos(pθ − λc = Labia + Labib + Laic + M cos(pθ + )if )if 2π 3 2π 3 λf = M cos(pθ)ia + M cos(pθ 2π 3 − )ib + M cos(pθ + ) + Lf if 2π 3 It is assumed that the ﬂux linkages are sinusoidal functions of rotor position. As it turns out, many electrical machines work best (by many criteria such as smoothness of torque production) 8 if this is the case, so that techniques have been developed to make those ﬂux linkages very nearly sinusoidal. We will see some of these techniques in later chapters of these notes. For the moment, we will simply assume these dependencies. In addition, we assume that the rotor is magnetically ’round’, which means the stator self inductances and the stator phase to phase mutual inductances are not functions of rotor position. Note that if the phase windings are identical (except for their If there are three uniformly spaced angular position), they will have identical self inductances. windings the phase-phase mutual inductances will all be the same. Now, this system can be simply described in terms of coenergy. With multiple excitation it is important to exercise some care in taking the coenergy integral (to ensure that it is taken over a valid path in the multi	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
multiple excitation it is important to exercise some care in taking the coenergy integral (to ensure that it is taken over a valid path in the multi-dimensional space). In our case there are actually ﬁve dimensions, but only four are important since we can position the rotor with all currents at zero so there is no contribution to coenergy from setting rotor position. Suppose the rotor is at some angle θ and that the four currents have values ia0, ib0, ic0 and if 0. One of many correct path integrals to take would be: W ′ m = + + + ia0 0 Z ib0 0 Z ic0 0 Z if 0 0 Z The result is: W ′ m = 1 2 Laiadia (Labia0 + Laib) dib (Labia0 + Labib0 + Laic) dic M cos(pθ)ia0 + M cos(pθ (cid:18) 2π 3 − )ib0 + M cos(pθ + )ic0 + Lf if 2π 3 dif (cid:19) L a i2 + i2 a0 2 b0 + ico + Lab (iaoib0 + iaoic0 + icoib0) (cid:16) +M if 0 (cid:17) ia0 cos(pθ) + i b0 cos(pθ (cid:18) 2π 3 2π ) + ic0 cos(pθ + ) 3 (cid:19) − 1 2 + Lf i f 0 2 Since there are no variations of the stator inductances with rotor position θ, torque is easily given by: Te = ∂W ′ m ∂θ = − pM if 0 ia0 sin(pθ) + ib0 sin(pθ (cid:18) 2 π 3 π 2 ) + ico sin(pθ + ) 3 (cid:19) − 3.2.1 Current Driven Synchronous Machine Now assume that we can drive this thing with currents: ia0 = Ia cos ωt ib0 = Ia cos ω t (cid:18) ic0 = Ia cos �	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
ia0 = Ia cos ωt ib0 = Ia cos ω t (cid:18) ic0 = Ia cos ω t + 2π − 3 2π 3 (cid:19) (cid:19) if 0 = If (cid:18) 9 and assume the rotor is turning at synchronous speed: pθ = ωt + δi Noting that cos x sin y = 1 sin(x 2 − y) + 1 sin(x + y), we ﬁnd the torque expression above to be: 2 Te = pM IaIf − (cid:18) + + 1 2 (cid:18) 1 2 1 2 sin δi + sin (2ωt + δi) sin δi + sin (cid:19) 2ωt + δi (cid:18) 4 π − 3 4π 3 (cid:19)(cid:19) sin δi + sin 2ωt + δi + 1 2 1 2 1 2 (cid:18) The sine functions on the left add and the ones on the right cancel, leaving: (cid:19)(cid:19) (cid:18) Te = 3 2 − pM IaIf sin δi And this is indeed one way of looking at a synchronous machine, which produces steady torque if the rotor speed and currents all agree on frequency. Torque is related to the current torque angle δi. As it turns out such machines are not generally run against current sources, but we will take up actual operation of such machines later. 4 Field Descriptions: Continuous Media While a basic understanding of electromechanical devices is possible using the lumped parameter approach and the principle of virtual work as described in the previous section, many phenomena in electric machines require a more detailed understanding which is aﬀorded by a continuum approach. In this section we consider a ﬁelds-based approach to energy ﬂow using Poynting’s Theorem and then a ﬁelds based description of forces using the Maxwell Stress Tensor. These techniques will both be useful in further analysis of what happens in electric machines. 4.1 Field Description of Energy Flow: Poyting’s Theorem Start with Faraday’s Law: and Ampere’s	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
happens in electric machines. 4.1 Field Description of Energy Flow: Poyting’s Theorem Start with Faraday’s Law: and Ampere’s Law: ~E = ~∂B − ∂t ∇ × ~H = J ~ ∇ × Multiplying the ﬁrst of these by H and the second by E and taking the diﬀerence: ~ ~ ~H ~E ~E − · ∇ × · ∇ × ~H = ∇ · ~E ~ H × (cid:16) (cid:17) = ~H − ∂B~ dt − · ~E J ~ · On the left of this expression is the divergence of electromagnetic energy ﬂow: ~ ~ S = E ~H × 10 ~ Here, S is the celebrated Poynting ﬂow which describes power in an electromagnetic ﬁeld sysstem. (The units of this quantity is watts per square meter in the International System). On ~ is rate of change of magnetic stored energy. The second the right hand side are two terms: H ~J looks a lot like power dissipation. We will discuss each of these in more detail. For the ~ term, E moment, however, note that the divergence theorem of vector calculus yields: ~ ∂B dt · · vo ume ∇ · Z l ~dv = S ~nda ~S · (cid:13) Z Z that is, the volume integral of the divergence of the Poynting energy ﬂow is the same as the Poynting energy ﬂow over the surface of the volume in question. This integral becomes: ~ S ~nda = · (cid:13) ZZ − volume Z ~ ~ ~ E J + H · ~∂B · ∂t ! dv which is simply a realization that the total energy ﬂow into a region of space is the same as the volume integral over that region of the rate of change of energy stored plus the term that looks like dissipation. Before we close this, note that, if there is motion of any material within the system, we can use the empirical expression for transformation of electric ﬁeld between observers moving with respect	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
, if there is motion of any material within the system, we can use the empirical expression for transformation of electric ﬁeld between observers moving with respect to each other. Here the ’primed’ frame is moving with respeect to the ’unprimed’ frame with the velocity ~v ′ ~E ~= E + ~v ~B × This transformation describes, for example, the motion of a charged particle such as an electron under the inﬂuence of both electric and magnetic ﬁelds. Now, if we assume that there is material motion in the system we are observing and if we assign ~v to be the velocity of that material, so that ~E′ is measured in a frame in which thre is no material motion (that is the frame of the material itself), the product of electric ﬁeld and current density becomes: ~E ~J = ′ ~E · ~B ~v − × · ~ ~ J = E ′ ~J · − ~B ~v × · ~ ~ J = E ′ ~J + ~v · ~J ~B × · (cid:16) (cid:16) In the last step we used the fact that in a scalar triple product the order of the scalar (dot) and vector (cross) products can be interchanged and that reversing the order of terms in a vector (cross) product simply changes the sign of that product. Now we have a ready interpretation for what we have calculated: (cid:16) (cid:17) (cid:17) (cid:17) If the ’primed’ coordinate system is actually the frame of material motion, ′ ~E · ~J = 1 σ \| ~J 2 \| which is easily seen to be dissipation and is positive deﬁnite if material conductivity σ is positive. The last term is obviously conversion of energy from electromagnetic to mechanical form: where we have now identiﬁed force density to be: ~v · ~J × (cid:16) ~B = ~v ~F · (cid:17) ~ ~ F = J ~B × 11 This is the Lorentz Force Law, which describes the interaction of current with magnetic ﬁeld to produce force. It is not,	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
This is the Lorentz Force Law, which describes the interaction of current with magnetic ﬁeld to produce force. It is not, however, the complete story of force production in electromechanical systems. As we learned earlier, changes in geometry which aﬀect magnetic stored energy can also produce force. Fortunately, a complete description of electromechanical force is possible using only magnetic ﬁelds and that is the topic of our next section. 4.2 Field Description of Forces: Maxwell Stress Tensor Forces of electromagnetic origin, because they are transferred by electric and magnetic ﬁelds, are the result of those ﬁelds and may be calculated once the ﬁelds are known. In fact, if a surface can be established that fully encases a material body, the force on that body can be shown to be the integral of force density, or traction over that surface. The traction τ derived by taking the cross product of surface current density and ﬂux density on the air-gap surface of a machine (above) actually makes sense in view of the empirically derived Lorentz Force Law: Given a (vector) current density and a (vector) ﬂux density. This is actually enough to describe the forces we see in many machines, but since electric machines have permeable magnetic material and since magnetic ﬁelds produce forces on permeable material even in the absence of macroscopic currents it is necessary to observe how force appears on such material. A suitable empirical expression for force density is: where H is the magnetic ﬁeld intensity and µ is the permeability. ~ ~ ~ F = J ~B 1 − 2 × ~H ~H (cid:16) · (cid:17) µ ∇ Now, note that current density is the curl of magnetic ﬁeld intensity, so that: And, since: f orce den it s y is: ~F = ~H ∇ × (cid:16) = µ ∇ × (cid:16) ~µH × ~H × (cid:17) − − 1 2 1 (cid:16) 2 (cid:16) · ~H H µ ~ ∇ (cid:17) ~H ~H µ · ∇ (cid:17	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
) · ~H H µ ~ ∇ (cid:17) ~H ~H µ · ∇ (cid:17) (cid:17) ~H ~H × (cid:17) ∇ × (cid:16) ~ = H ~ H (cid:16) · ∇ (cid:17) ~H 1 − 2 ∇ ~F = µ = µ ~H ~H ~H (cid:16) ~H (cid:16) · ∇ · ∇ (cid:17) (cid:17) 1 2 µ − − ∇ (cid:18) ∇ 1 2 ~H (cid:16) µ · ~H (cid:16) ~ H (cid:17) ~H · − (cid:17)(cid:19) ~H ~H · (cid:17) ~ H (cid:16) ~H · (cid:17) µ ∇ (cid:16) 1 2 This expression can be written by components: the component of force in the i’th dimension is: Fi = µ H k X (cid:18) k The ﬁrst term can be written as: ∂ 1 ∂xk (cid:19) − ∂xi 2 Hi ∂ µ H k X 2 k ! Hk ∂ ∂xk (cid:19) µ X (cid:18) k Hi = k X µHkHi Hi − ∂ ∂xk µHk k X ∂ xk ∂ 12 The last term in this expression is easily shown to be divergence of magnetic ﬂux density, which is zero: ~B = ∇ · k X ∂ ∂xk µHk = 0 Using this, we can write force density in a more compact form as: Fk = ∂ ∂xi µH iHk µ 2 δ ik − H 2 n ! n X where we have used the Kroneker delta δik = 1 if i = k, 0	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
H 2 n ! n X where we have used the Kroneker delta δik = 1 if i = k, 0 otherwise. Note that this force density is in the form of the divergence of a tensor: Fk = ∂ ∂xi Tik or ∇ · In this case, force on some object that can be surrounded by a closed surface can be found by ~F = T using the divergence theorem: ~f = vol Z ~F dv = or, if we note surface traction to be τi = total force in direction i is just: vol ∇ · T dv = T ~nda · (cid:13) Z Z Z k Tiknk , where n is the surface normal vector, then the P ~f = s I τida = Tiknkda I X k The interpretation of all of this is less diﬃcult than the notation suggests. This ﬁeld description of forces gives us a simple picture of surface traction, the force per unit area on a surface. If we just integrate this traction over the area of some body we get the whole force on the body. Note one more thing about this notation. Sometimes when subscripts are repeated as they are here the summation symbol is omitted. Thus we would write τi = k Tiknk = Tiknk. 4.3 Example: Linear Induction Machine P Figure 7 shows a highly simpliﬁed picture of a single sided linear induction motor. This is not how most linear induction machines are actually built, but it is possible to show through symmetry arguments that the analysis we can carry out here is actually valid for other machines of this class. This machine consists of a stator (the upper surface) which is represented as a surface current on the surface of a highly permeable region. The moving element consists of a thin layer of conducting material on the surface of a highly permeable region. The moving element (or ’shuttle’) has a velocity u with respect to the stator and that motion is in the x direction. The stator surface current density is assumed to be: Kz = Re K ej(ωt−kx) n z o 13 K z g µ µ (cid:1)(cid:1)(cid:1)(	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0) (cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) y u x σ s K s Figure 7: Simple Model of single sided linear induction machine Note that we are ignoring some important eﬀects, such as those arising from ﬁnite length of the stator and of the shuttle. Such eﬀects can be quite important, but we will leave those until later, as they are what make linear motors interesting. Viewed from the shuttle for which the dimension in the direction of motion is x′ ut′, the x − − relative frequency is: ωt kx = (ω ku) t ′ kx = ωst ′ kx − Now, since the shuttle surface can support a surface current and is excited by magnetic ﬁelds which are in turn excited by the stator currents, it is reasonable to assume that the form of rotor current is the same as that of the stator: − − − K = Re s K ej(ωst s ′−kx ) Ampere’s Law is, in this situation: n o g ∂Hy ∂x = Kz + Ks which is, in complex amplitudes: The y- component of Faraday’s Law is, assuming the problem is uniform in the z- direction: H y = z + K	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
amplitudes: The y- component of Faraday’s Law is, assuming the problem is uniform in the z- direction: H y = z + K s K jkg − jω B s ′ y = jkEz − or ω s µ0H k A bit of algebraic manipulation yields expressions for the complex amplitudes of rotor surface ′ z = − E y current and gap magnetic ﬁeld: Ks = H y = σ j µ0ωs s k2g − 1 + j mu0ωsσs Kz k2g j K kg 1 + j mu0ωsσs z k2g 14 To ﬁnd surface traction, the Maxwell Stress Tensor can be evaluated at a surface just below the stator (on this surface the x- directed magnetic ﬁeld is simply H x = Kz. Thus the traction is τx = Txy = µ0HxHy and the average of this is: This is: < τx >= Re H µ0 2 xH y∗ o 2 0ωsσs k2g µ n K z\| \| µ0 1 2 kg 1 + Now, if we consider electromagnetic power ﬂow (Poynting(cid:17)’s Theorem): in the y- direction: < τx >= µ0ωsσs k2g (cid:16) 2 And since in the frame of the shuttle E < S ′ y >= 1 ω µ s 0 2 k kg − 2 K = z \| \| − ωs k < τ > x 2 ′ z = ωs Sy = EzHx − k µ0H y 0 sσs k2g 1 + µ0ωsσs g (cid:16) µ ω k2 (cid:17) Similarly, evaluated in the frame of the stator: ω k < Sy >= − < τ x > − This shows what we already suspected: the electromagnetic power ﬂow from the stator is the force density on the shuttle times the wave velocity. The electromagnetic	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
shows what we already suspected: the electromagnetic power ﬂow from the stator is the force density on the shuttle times the wave velocity. The electromagnetic ower ﬂow into the shuttle is the same force density times the ’slip’ velocity. The diﬀerence between these two is the power converted to mechanical form and it is the force density times the shuttle velocity. 4.4 Rotating Machines The use of this formulation in rotating machines is a bit tricky because, at lest formally, directional vectors must have constant identity if an integral of forces is to become a total force. In cylindrical coordinates, of course, the directional vectors are not of constant identity. However, with care and understanding of the direction of traction and how it is integrated we can make use of the MST approach in rotating electric machines. Now, if we go back to the case of a circular cylinder and are interested in torque, it is pretty clear that we can compute the circumferential force by noting that the normal vector to the cylinder is just the radial unit vector, and then the circumferential traction must simply be: τθ = µ0HrHθ Assuming that there are no ﬂuxes inside the surface of the rotor, simply integrating this over the surface gives azimuthal force. In principal this is the same as surrounding the surface of the rotor by a continuum of inﬁnitely small boxes, one surface just outside the rotor and with a normal facing outward, the other surface just inside with normal facing inward. (Of course the MST is zero on this inner surface). Then multiplying by radius (moment arm) gives torque. The last step is to note that, if the rotor is made of highly permeable material, the azimuthal magnetic ﬁeld just outside the rotor is equal to surface current density. 15 5 Generalization to Continuous Media Now, consider a system with not just a multiplicity of circuits but a continuum of current-carrying paths. In that case we could identify the co-energy as: W ′ m = ~ λ(~a)dJ d~a · area Z where that area is chosen to cut all of the current carrying conductors. This area can be picked to be perpedicular to each of the current ﬁlaments since the divergence of current is zero. The ﬂux λ	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
can be picked to be perpedicular to each of the current ﬁlaments since the divergence of current is zero. The ﬂux λ is calculated over a path that coincides with each current ﬁlament (such paths exist since current has zero divergence). Then the ﬂux is: Z Now, if we use the vector potential A for which the magnetic ﬂux density is: ~ λ(~a) = ~ B · d~n Z ∇ × the ﬂux linked by any one of the current ﬁlaments is: ~B = ~A where dℓ is the path around the current ﬁlament. This implies directly that the coenergy is: ~ λ(~a) = ~A ~ dℓ · I W ′ m = area ZJ I Z ~ ~ ~ A dℓdJ d~a · · Now: it is possible to make dℓ coincide with d~a and be parallel to the current ﬁlaments, so that: ~ 5.1 Permanent Magnets W ′ m = ~dJdv ~A · vol Z Permanent magnets are becoming an even more important element in electric machine systems. Often systems with permanent magnets are approached in a relatively ad-hoc way, made equivalent to a current that produces the same MMF as the magnet itself. The constitutive relationship for a permanent magnet relates the magnetic ﬂux density B to ~ magnetic ﬁeld H and the property of the magnet itself, the magnetization M . ~ ~ ~B = µ0 ~H + M ~ (cid:16) Now, the eﬀect of the magnetization is to act as if there were a current (called an amperian current) with density: (cid:17) ∗ ~J = ∇ × ~M Note that this amperian current “acts” just like ordinary current in making magnetic ﬂux density. Magnetic co-energy is: W ′ m = ~A · ∇ × ~dM dv l vo Z 16 Next, note the vector identity ~ C × ∇ · (cid:16) ~ D ~ = D (cid:17) · ∇ × (cid:	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
~ C × ∇ · (cid:16) ~ D ~ = D (cid:17) · ∇ × (cid:16) ~ C · − ~ C (cid:17) ′ Wm = Then, noting that B = ~ ∇ × −∇ · vol Z ~ A: ~ ~ A dM dv + × (cid:16) (cid:17) vol Z ∇ × (cid:16) · (cid:17) ~D Now, ∇ × (cid:16) ~ A (cid:17) ~ dM dv W ′ m = − (cid:13) Z Z ~dM d~s + ~A × ~dM dv ~B · vol Z The ﬁrst of these integrals (closed surface) vanishes if it is taken over a surface just outside the magnet, where M is zero. Thus the magnetic co-energy in a system with only a permanent magnet source is ~ W ′ m = ~ dM dv ~ B · v ol Z Adding current carrying coils to such a system is done in the obvious way. 17 MIT OpenCourseWare http://ocw.mit.edu 6.685 Electric Machines Fall 2013 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.	https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf
8.022 (E&M) – Lecture 6 Topics: (cid:132) More on capacitors (cid:132) Mini-review of electrostatics (cid:132) (almost) all you need to know for Quiz 1 Last time… (cid:132) Capacitor: (cid:132) System of charged conductors (cid:132) Capacitance: C = Q V (cid:132) It depends only on geometry +Q + + + + + + -- -Q - - - - (cid:132) Energy stored in capacitor: (cid:132) In agreement with energy associated with electric field U = 2 Q C 2 = 1 2 2 CV (cid:132) Let’s now apply what we have learned… G. Sciolla – MIT 8.022 – Lecture 6 2 1 Wimshurst machine and Leyden Jars (E1) (cid:132) A Wimshurst machine is used to charge 2 Leyden Jars (cid:132) Leyden Jars are simple cylindrical capacitors Insulator Outer conductor Inner conductor (cid:132) What happens when we connect the outer and the outer surface? (cid:132) Why? G. Sciolla – MIT 8.022 – Lecture 6 3 Dissectible Leyden Jar (E2) (cid:132) A Wimshurst machine is used to charge a Leyden Jar (cid:132) Where is the charge stored? (cid:132) On the conductors? (cid:132) On the dielectric? (cid:132) Take apart capacitor and short conductors (cid:132) Nothing happens! (cid:132) Now reassemble it (cid:132) Bang! (cid:132) Why? (cid:132) Because it’s “easier” for the charges to stay on dielectric when we take conductors apart or energy stored would have to change: U=Q2/2C, and moving plates away C would decrease (cid:198) U increase G. Sciolla – MIT 8.022 – Lecture 6 4 2 Capacitors and dielectrics (cid:132) Parallel plates capacitor: C = Q Q = Ed V = A dπ 4 (cid:132) Add a	https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf
:132) Parallel plates capacitor: C = Q Q = Ed V = A dπ 4 (cid:132) Add a dielectric between the plates: (cid:132) Dielectric’s molecules are not spherically symmetric (cid:132) Electric charges are not free to move (cid:198) E will pull + and – charges apart and orient them // E + + + + + + + + + + E=4πσ _ + _ + _ + _ + _ + _ + _ + E (cid:132) Edielectric is opposite to Ecapacitor (cid:132) Given Q (cid:198) V decreases (cid:132) Given V (cid:198) Q increases } (cid:198) C increased! G. Sciolla – MIT 8.022 – Lecture 6 - - - - - - - - - - - - 5 Energy is stored in capacitors (E6) (cid:132) A 100 µF oil filled capacitor is charged to 4KV (cid:132) What happens if we discharge it thought a 12” long iron wire? + + + + + Oil - - - - - (cid:132) How much energy is stored in the capacitor? (cid:132) U= ½ CV2 = 800 J Big! (cid:132) Resistance of iron wire: very small, but >> than the rest of the circuit (cid:198) All the energy is dumped on the wire in a small time (cid:198) Huge currents! (cid:198) Huge temperatures! (cid:198) The wire will explode! G. Sciolla – MIT 8.022 – Lecture 6 6 3 Capacitors in series (cid:132) Let’s connect 2 capacitors C1 and C2 in the following way: Q1 Q2 A B V V1 V2 C ? V (cid:132) What is the total capacitance C of the new system? V V + 1 2 Q Q = 1 1 V = C Q = V = Q 2 = 1 V V + Q 2 = 1 C 1 + 1 C 2	https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf
= V = Q 2 = 1 V V + Q 2 = 1 C 1 + 1 C 2 B - + - + - + - + - + - +- This is 1 conductor that starts electrically neutral (cid:198) Q1=Q2 G. Sciolla – MIT 1 − C ⎞ ⎟ ⎠ i 8.022 – Lecture 6 ⎛ = ⎜ ⎝ 1 C i ∑ Capacitors in parallel (cid:132) Let’s connect 2 capacitors C1 and C2 in the following way: Q1 Q2 V1 V2 V ? V (cid:132) What is the total capacitance C of the new system? 1 + = = V V V = 1 2 Q Q Q 2 Q Q + 2 V C = 1 = Q Q 2 1 + V V 2 1 = C C + 1 2 (cid:198) C 7 C i i N = = ∑ i 1 = G. Sciolla – MIT 8.022 – Lecture 6 8 4 Application (cid:132) Why are capacitors useful? (cid:132) …among other things… (cid:132) They can store large amount of energy and release it in very short time (cid:132) Energy stored: U= ½ CV2 (cid:132) The larger the capacitance, the larger the energy stored at a given V (cid:132) How to increase the capacitance? (cid:132) Modify geometry (cid:132) For parallel plates capacitors C=A/(4πd): increase A or decrease d (cid:132) Add a dielectric in between the plates (cid:132) Add capacitors in parallel G. Sciolla – MIT 8.022 – Lecture 6 9 Bank of capacitors (E7) (cid:132) Bank of 12 x 80 µF capacitors is parallel V ------- ------- ++++++ ++++++ … ------- ++++++ 80µF 60W (cid:132) Total capacitance: 960 µF (cid:132)	https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf
++++++ ++++++ … ------- ++++++ 80µF 60W (cid:132) Total capacitance: 960 µF (cid:132) Discharged on a 60 W light bulb when capacitors are charged at: (cid:132) V = 100 V, 200 V, V = 300 V (cid:132) What happens? (cid:132) (cid:132) Energy stored in capacitor is U= ½ CV2 (cid:198) V = V0: 2xV0 : 3xV0 (cid:198) U = U0 : 4xU0 : 9xU0 R is the same (cid:198) time of discharge will not change with V The power will increase by a factor 9! (P=RI2 and I=V/R) (cid:132) (cid:132) Will the bulb survive? (cid:132) Remember: light bulb designed for 120 V… G. Sciolla – MIT 8.022 – Lecture 6 10 5 Review of Electrostatics for Quiz 1 Disclaimer: (cid:132) Can we review all of the electrostatics in less than 1 hour? (cid:132) No, but we will try anyway… (cid:132) Only main concepts will be reviewed (cid:132) Review main formulae and tricks to solve the various problems (cid:132) No time for examples (cid:132) Go back to recitations notes or Psets and solve problems again G. Sciolla – MIT 8.022 – Lecture 6 11 The very basic: Coulomb’s law (cid:71) F 2 = q q 1 r 2 1 \| 2 2 \| ˆ r 2 1 where F2 is the force that the charge q2 feels due to q1 NB: this is in principle the only thing you have to remember: all the rest follows from this an the superposition principle G. Sciolla – MIT 8.022 – Lecture 6 12 6 The very basic: Superposition principle qN q5 q2 q3 q1 q4 qi Q r V Q (cid:71) F Q i N = = ∑ i = 1 q Q	https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf
Q r V Q (cid:71) F Q i N = = ∑ i = 1 q Q i 2 r \| i \| ˆ r i (cid:71) QF = ∫ V dq Q 2 \|r\| ˆ r = ∫ V ρ dV Q 2 \|r\| ˆ r G. Sciolla – MIT 8.022 – Lecture 6 13 The Importance of Superposition Extremely important because it allows us to transform complicated problems into sum of small, simple problems that we know how to solve. Example: + + + + + Empty + Empty + Empty + + + Calculate force F exerted by this distribution of charges on the test charge q q G. Sciolla – MIT 8.022 – Lecture 6 14 7 Electric Field and Electric Potential (cid:132) Solving problems in terms of Fcoulomb is not always convenient (cid:132) F depends on probe charge q (cid:132) We get rid of this dependence introducing the Electric Field (cid:71) qF q Q \| r (cid:132) Advantages and disadvantages of E (cid:71) E = = ˆ r \| 2 (cid:132) E describes the properties of space due to the presence of charge Q ☺ (cid:132) It’s a vector (cid:198) hard integrals when applying superposition… (cid:47) (cid:132) Introduce Electric Potential φ (cid:132) φ(P) is the work done to move a unit charge from infinity to P(x,y,z) ( φ x y z , , ) = − ∫ P ∞ (cid:74)(cid:71) (cid:71) i E ds NB: true only when φ(inf)=0 (cid:132) Advantages: superposition still holds but simpler calculation (scalar) ☺ G. Sciolla – MIT 8.022 – Lecture 6 15 Energy associated with E (cid:132) Moving charges in E requires work: 2 W − > = − 2 1 ∫ 1 (cid:71) F C (cid:71) ds • w	https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf
W − > = − 2 1 ∫ 1 (cid:71) F C (cid:71) ds • w here F C oulom b = ˆ Q qr 2 r y 1 2 P1 P2 3 x (cid:132) NB: integral independent of path: force conservative! (cid:132) Assembling a system of charges costs energy. This is the energy stored in the electric field: U = 1 2 ∫ Volume with charges ρφ dV = 2 E 8 π dV ∫ Entire space G. Sciolla – MIT 8.022 – Lecture 6 16 8 Electrostatics problems (cid:132) In electrostatics there are 3 different ways of describing a problem: ρ(x,y,z) Ε(x,y,z) φ(x,y,z) (cid:132) Solving most problem consists in going from one formulation to another. All you need to know is: how? G. Sciolla – MIT 8.022 – Lecture 6 17 From ρ (cid:198) E (cid:132) General case: (cid:132) For a point charge: (cid:71) E (cid:132) Superposition principle: (cid:132) Special cases: q r \| = = (cid:71) E 2 ˆ r \| ∫ V (cid:71) d E d q r \| 2 ˆ r = ∫ V \| Solving this integral may not be easy… (cid:132) Look for symmetry and thank Mr. Gauss who solved the integrals for you (cid:132) Gauss’s Law: (cid:118) ∫ (cid:132) N.B.: S Qπ 4 Φ =(cid:71) E (cid:71) (cid:71) i E d A = enc ∫ 4 π ρ V d V S +Q S1 (cid:132) Gauss’s law is always true but not always useful: Symmetry is needed! (cid:132) Main step: choose the “right” gaussian surface so that E is constant on the	https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf
not always useful: Symmetry is needed! (cid:132) Main step: choose the “right” gaussian surface so that E is constant on the surface of integration G. Sciolla – MIT 8.022 – Lecture 6 18 9 From ρ (cid:198) φ (cid:132) General case: (cid:132) For a point charge: (cid:132) Superposition principle: φ = q r φ = ∫ V NB: implicit hypothesis: φ(infinity)=0 d q r The problem is simpler than for E (only scalars involved) but not trivial… (cid:132) Special cases: (cid:132) If symmetry allows, use Gauss’s law to extract E and then integrate E to get φ: φ φ− 1 2 2 = − ∫ 1 (cid:74)(cid:71) (cid:71) i E ds (cid:132) N.B.: The force is conservative (cid:198) the result is the same for any path, but choosing a simple one makes your life much easier…. G. Sciolla – MIT 8.022 – Lecture 6 19 From φ to E and ρ Easy! No integration needed! (cid:132) From φ to E (cid:71) E = − ∇ φ (cid:132) One derivative is all it takes but… make sure you choose the best coordinate system (cid:132) You will not loose points but you will waste time… (cid:132) From φ to ρ (cid:132) Poisson tells you how to get from potential to charge distributions directly: 2 4 ∇ = − φ πρ (cid:132) Uncomfortable with Laplacian? Get there in 2 steps: (cid:71) E (cid:132) First calculate E: (cid:132) The use differential form of Gauss’s law: = − ∇ φ (cid:71) i E πρ 4 ∇ = G. Sciolla – MIT 8.022 – Lecture 6 20 10 Thoughts about φ and E (cid:132) The potential φ is always continuous (cid:132) E is not always continuous: it can “jump” (cid:132) When we have surface charge distributions	https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf
φ is always continuous (cid:132) E is not always continuous: it can “jump” (cid:132) When we have surface charge distributions (cid:132) Remember problem #1 in Pset 2 (cid:198) When solving problems always check for consistency! G. Sciolla – MIT 8.022 – Lecture 6 21 Summary ρ(x,y,z) 2 φ πρ 4 ∇ = − G a uss’s la w (int) (cid:71) i E πρ 4 ∇ = Ε(x,y,z) 2 φ φ− 1 (cid:71) E (cid:74)(cid:71) (cid:71) i E ds 2 = −∫ 1 = −∇ φ φ = ∫ V d q r φ(x,y,z) G. Sciolla – MIT 8.022 – Lecture 6 22 11 Conductors (cid:132) Properties: (cid:132) Surface of conductors are equipotential (cid:132) E (field lines) always perpendicular to the surface (cid:132) Einside=0 (cid:132) Esurface=4πσ (cid:132) What’s the most useful info? (cid:132) Einside=0 because it comes handy in conjunction with Gauss’s law to solve problems of charge distributions inside conductors. + (cid:132) Example: concentric cylindrical shells (cid:132) Charge +Q deposited in inner shell (cid:132) No charge deposited on external shell (cid:132) What is E between the 2 shells? (cid:132) - Q induced on inner surface of inner cylinder (cid:132) +Q induced on outer surface of outer cylinder G. Sciolla – MIT 8.022 – Lecture 6 + + - + + - - + - + + + + - + + - - + + 23 E due to Charges and Conductors (cid:132) How to find E created by charges near conductors? (cid:132) Uniqueness theorem: (cid:132) A solution that satisfies boundary conditions is THE solution (cid:132) Be creative and think of distribution of point charges that will create the same filed lines: (cid:132) Example: Method of images + +	https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf
Be creative and think of distribution of point charges that will create the same filed lines: (cid:132) Example: Method of images + + - - - - - - - - - - - - - - - - - - - G. Sciolla – MIT 8.022 – Lecture 6 24 12 Capacitors (cid:132) Capacitance have capacitance C (cid:132) Two oppositely charged conductors kept at a potential difference V will Q V C = (cid:132) NB: capacitance depends only on the geometry! (cid:132) Energy stored in capacitor U = 2 Q C 2 = 1 2 2 CV (cid:132) What should you remember? (cid:132) Parallel plate capacitor: very well (cid:132) Be able to derive the other standard geometries G. Sciolla – MIT 8.022 – Lecture 6 25 Conclusion (cid:132) Material for Quiz #1: (cid:132) Up to this lecture (Purcell chapters 1/2/3) (cid:132) Next lecture: (cid:132) Charges in motion: currents (cid:132) NB: currents are not included in Quiz 1! G. Sciolla – MIT 8.022 – Lecture 6 26 13	https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf
2 General results of representation theory 2.1 Subrepresentations in semisimple representations Let A be an algebra. Deﬁnition 2.1. A semisimple (or completely reducible) representation of A is a direct sum of irreducible representations. Example. Let V be an irreducible representation of A of dimension n. Then Y = End(V ), with action of A by left multiplication, is a semisimple representation of A, isomorphic to nV (the direct sum of n copies of V ). Indeed, any basis v1, ..., vn of V gives rise to an isomorphism of representations End(V ) nV , given by x (xv1, ..., xvn). ⊃ ⊃ Remark. Note that by Schur’s lemma, any semisimple representation V of A is canonically X, where X runs over all irreducible representations of A. Indeed, Hom(X, V ), x identiﬁed with we have a natural map f : and it is easy to verify that this map is an isomorphism. � �X Hom(X, V ) �X HomA(X, V ) V , given by g g(x), x X, g X ⊃ ⊃ � � � � We’ll see now how Schur’s lemma allows us to classify subrepresentations in ﬁnite dimensional semisimple representations. i Proposition 2.2. Let Vi, 1 ∗ representations of A, and W be a subrepresentation of V = m be irreducible ﬁnite dimensional pairwise nonisomorphic m i=1niVi. Then W is isomorphic to niVi given � by multiplication of a row vector of elements of	https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0721b66f53c3c196ce86d6d867514442_MIT18_712F10_ch2.pdf
is isomorphic to niVi given � by multiplication of a row vector of elements of Vi (of length ri) by a certain ri-by-ni matrix Xi with linearly independent rows: θ(v1, ..., vri ) = (v1, ..., vri )Xi. V is a direct sum of inclusions θi : riVi ni, and the inclusion θ : W m i=1riVi, ri ⊃ ⊃ ∗ ∗ � by induction in n := m i=1 ni. The base of induction (n = 1) is clear. To perform Proof. The proof is the induction step, let us assume that W is nonzero, and ﬁx an irreducible subrepresentation W . Such P exists (Problem 1.20). 2 Now, by Schur’s lemma, P is isomorphic to Vi for some i, P V factors through niVi, and upon identiﬁcation of P with Vi is given and the inclusion θ by the formula v \|P : P (vq1, ..., vqni ), where ql k are not all zero. ⎨ ⊃ → �⊃ � ⊃ Now note that the group Gi = GLni (k) of invertible ni-by-ni matrices over k acts on niVi (v1, ..., vni )gi (and by the identity on njVj , j = i), and therefore acts on the by (v1, ..., vni ) set of subrepresentations of V , preserving the property we need to establish: namely, under the action of gi, the matrix Xi goes to Xigi, while Xj , j = i don’t change.	https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0721b66f53c3c196ce86d6d867514442_MIT18_712F10_ch2.pdf
the matrix Xi goes to Xigi, while Xj , j = i don’t change. Take gi Gi such that (q1, ..., qni )gi = (1, 0, ..., 0). Then W gi contains the ﬁrst summand Vi of niVi (namely, it is P gi), hence W gi = Vi nmVm is the kernel of the projection − of W gi to the ﬁrst summand Vi along the other summands. Thus the required statement follows from the induction assumption. W �, where W � → n1V1 1)Vi (ni ... ... � � � � � � Remark 2.3. In Proposition 2.2, it is not important that k is algebraically closed, nor it matters that V is ﬁnite dimensional. If these assumptions are dropped, the only change needed is that the entries of the matrix Xi are no longer in k but in Di = EndA(Vi), which is, as we know, a division algebra. The proof of this generalized version of Proposition 2.2 is the same as before (check it!). 2Another proof of the existence of P , which does not use the ﬁnite dimensionality of V , is by induction in n. Namely, if W itself is not irreducible, let K be the kernel of the projection of W to the ﬁrst summand V1. Then K is a subrepresentation of (n1 − 1)V1 � ... � nmVm, which is nonzero since W is not irreducible, so K contains an irreducible subrepresentation by the induction assumption. 2.2 The density theorem Let A be an algebra over an algebraically	https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0721b66f53c3c196ce86d6d867514442_MIT18_712F10_ch2.pdf
subrepresentation by the induction assumption. 2.2 The density theorem Let A be an algebra over an algebraically closed ﬁeld k. Corollary 2.4. Let V be an irreducible ﬁnite dimensional representation of A, and v1, ..., vn V there exists an element a be any linearly independent vectors. Then for any w1, ..., wn such that avi = wi. � V A � � (av1, ..., avn) is a Proof. Assume the contrary. Then the image of the map A proper subrepresentation, so by Proposition 2.2 it corresponds to an r-by-n matrix X, r < n. Thus, vn). Let taking a = 1, we see that there exist vectors u1, ..., ur (q1, ..., qn) be a nonzero ivi = T qivi = 0 - a contradiction with the linear independence of (u vi. V such that (u1, ..., ur)X = (v1, ..., that X(q1, ..., qn)T = 0 (it exists because r < n). Then vector such 1, ..., ur )X(q1, ..., qn) = 0, i.e. nV given by a ⎨ ⊃ ⊃ � q ⎨ Theorem 2.5. (the Density Theorem). (i) Let V be an irreducible ﬁnite dimensional representation of A. Then the map δ : A EndV is surjective. (ii) Let V = V1 ... representations of A. Then � � ⊃ Vr, where Vi are irreducible pairwise nonisomorphic ﬁnite dimensional ⊃ �i=1 End(Vi) is surjective. r i=1δi : A the map � r Proof	https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0721b66f53c3c196ce86d6d867514442_MIT18_712F10_ch2.pdf
=1 End(Vi) is surjective. r i=1δi : A the map � r Proof. (i) Let B be the image of A in End(V ). We want to show that B = End(V ). Let c v1, ..., vn be a basis of V , and wi = cvi. By Corollary 2.4, there exists a B, and we are done. Then a maps to c, so c End(V ), A such that avi = wi. � � � i be the image of A in (ii) Let B a representation A, Then by Proposition 2.2, B = follows. of � End(V i), and B be the image of A in r i=1 End(Vi) is semisimple: it is isomorphic to �i=1 End(Vi). Recall that as r i=1diVi, where di = dim Vi. � �iBi. On the other hand, (i) implies that Bi = End(Vi). Thus (ii) r 2.3 Representations of direct sums of matrix algebras In this section we consider representations of algebras A = r i Matdi (k) for any ﬁeld k. i=1 Matdi (k). Then the irreducible� representations of A are V1 = r = k r , and any ﬁnite dimensional representation of A is a direct sum of copies of = � Theorem 2.6. Let A kd1 , . . . , V V1, . . . , Vr. d In order to prove Theorem 2.6, we shall need the notion of a dual representation. Deﬁnition 2.7. (Dual representation) Let V be a representation of any algebra A. Then the	https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0721b66f53c3c196ce86d6d867514442_MIT18_712F10_ch2.pdf
7. (Dual representation) Let V be a representation of any algebra A. Then the dual representation V ⊕ is the representation of the opposite algebra Aop (or, equivalently, right A-module) with the action a)(v) := f (av). (f · Proof of Theorem 2.6. First, the given representations are clearly irreducible, as for any v = 0, w � A such that av = w. Next, let X be an n-dimensional representation of Vi, there exists a (Matdi (k)) ∪= Matdi (k) with A. Then, X ⊕ isomorphism �(X) = X T , as (BC)T = C T BT . Thus, A =∪ Aop and X ⊕ may be viewed as an n-dimensional representation of A. Deﬁne representation of Aop. But is an n-dimensional � op θ : A � · · · � n copies �� A X ⊕ −⊃ � by θ(a1, . . . , an) = a1y1 + + anyn · · · is a basis of X ⊕. θ is clearly surjective, as k A. Thus, the dual map θ⊕ : X An where yi} { is injective. But A = An as representations of A (check it!). Hence, Im θ⊕ ∪= X is a subrepresen tation of An �i=1ndiVi, as a representation of A. Hence by Proposition 2.2, X . Next, Matdi (k) = diVi, so A = r i=1diVi, A = n ⊃ →	https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0721b66f53c3c196ce86d6d867514442_MIT18_712F10_ch2.pdf

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