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free R-module for each n. Then the cross product map H k(X; R) ⊗ H (Y ; R) × H n(X × Y ; R) k+=n is an isomorphism for every n, for every finite cell complex X. It follows from the five lemma that the same holds if we have a relative complex (Y, A) instead of just Y. For convenience, we will write H ∗(X; R)⊗H ∗(Y ; R) f... |
which gives us a map F n(X, A) → Gn(X, A). It is immediate Gn has a long exact sequence associated to (X, A) given by the usual long exact sequence of (X × Y, A × Y ). We would like to say F has a long exact sequence as well, and this is where our hypothesis comes in. If H ∗(Y ; R) is a free R-module, then we can take... |
F n(X, X ) → Gn(X, X ) is an isomorphism. We now notice that F ∗(−) and G∗(−) have excision. Since (X, X ) is a good pair, we have a commutative square F ∗(Dn, ∂Dn) × G∗(Dn, ∂Dn) F ∗(X, X ) × G∗(X, X ) ∼ ∼ So we now only need the left-hand map to be an isomorphism. We look at the long exact sequence for (Dn, ∂Dn)! F ∗... |
H ∗(S1; Z)⊗n, where T n = (S1)n is the n-torus, and this is an isomorphism of rings. So this is H ∗(T n, Z) ∼= Z[x1, · · ·, xn]/(x2 i, xixj + xjxi), using the fact that xi, xj have degree 1 so anti-commute. Note that this has an interesting cup product! This ring is known as the exterior algebra in n generators. Examp... |
relate to the usual homology and cohomology theory. Proof. Let Cn be Cn(X; R) and Zn ⊆ Cn be the cycles and Bn ⊆ Zn the boundaries. Then there is a short exact sequence 0 Zn i Cn g Bn−1 0, and Bn−1 ≤ Cn−1 is a submodule of a free R-module, and is free, since R is a PID. So by picking a basis, we can find a map s : Bn−1... |
with first two terms equal, the last terms have to be equal as well. The cohomology version is similar. 55 10 Vector bundles III Algebraic Topology 10 Vector bundles 10.1 Vector bundles We now end the series of random topics, and work on a more focused topic. We are going to look at vector bundles. Intuitively, a vecto... |
, e) ∈ Y × E : f (y) = π(e)}. This has a map f ∗π : f ∗E → Y given by projecting to the first coordinate. The vector space structure on each fiber is given by the identification (f ∗E)y = Ef (y). It is a little exercise in topology to show that the local trivializations of π : E → X induce local trivializations of f ∗π : ... |
” vector bundle, as we will see later. Example (Grassmannian manifold). We let X = Grk(Rn) = {k-dimensional linear subgroups of Rn}. To topologize this space, we pick a fixed V ∈ Grk(Rn). Then any k-dimensional subspace can be obtained by applying some linear map to V. So we obtain a surjection GLn(R) → Grk(Rn) M → M (V... |
T M is a subbundle of i∗T N. Note that we cannot say T M is a smooth subbundle of T N, since they have different base space, and thus cannot be compared without pulling back. The normal bundle of M in N is νM ⊆N = i∗T N T M. Here is a theorem we have to take on faith, because proving it will require some differential ge... |
up using partitions of unity. Let {Uα}α∈I be an open cover of X with local trivializations ϕα : E|Uα → Uα × Rd. The inner product on Rd then gives us an inner product on E|Uα, say ·, · α. We let λα be a partition of unity associated to {Uα}. Then for u ⊗ v ∈ E ⊗ E, we define u, v = λα(π(u))u, vα. α∈I Now if π(u) = π(v)... |
then the first coordinate distinguishes them. If they are in the same fiber, then there is some Ui with λi(π(u)) = 0. Then looking at the ith coordinate gives us distinguishes them. This then exhibits E as a subbundle of X × Rn. Corollary. Let π : E → X be a vector bundle over a compact Hausdorff space. Then there is som... |
Grd(R∞), as there can be multiple embeddings of X into the trivial bundle X × RN. Indeed, if we wiggle the embedding a bit, then we can get a new bundle. So we don’t have a coorespondence between vector bundles π : E → X and maps fπ : X → Grd(R∞). The next best thing we can hope for is that the “wiggle the embedding a... |
E# x ; R), where E# x = Ex \ {0}. We know Ex is a d-dimensional vector space. So we can choose an isomorphism Ex → Rd. So after making this choice, we know that H i(Ex, E# x ; R) ∼= H i(Rd, Rd \ {0}; R) = R i = d 0 otherwise However, there is no canonical generator of H d(Ex, E# we had to pick an isomorphism Ex ∼= Rd.... |
. has a positive determinant on each fiber, then E is orientable for any R. Note that we don’t have to check the determinant at each point on Uα ∩ Uβ. By continuity, we only have to check it for one point. Proof. Choose a generator u ∈ H d(Rd, Rd \ {0}; R). Then for x ∈ Uα, we define εx by pulling back u along Ex E|Uα ϕα... |
(iii), since H i(X; R) = 0 for i < 0. Before we go on and prove this, we talk about why it is useful. Definition (Euler class). Let π : E → X be a vector bundle. We define the Euler class e(E) ∈ H d(X; R) by the image of uE under the composition H d(E, E#; R) H d(E; R) s∗ 0 H d(X; R). 62 10 Vector bundles III Algebraic ... |
∗ 0 s∗ H d(X; R) Since s and s0 are homotopic, the diagram commutes. Also, the top row is exact. So uE ∈ H d(E, E#; R) gets sent along the top row to 0 ∈ H d(E#; R), and thus s∗ sends it to 0 ∈ H d(X; R). But the image in H d(X; R) is exactly the Euler class. So the Euler class vanishes. Now cohomology is not only a bu... |
(c d e(E)). So e(E) is precisely the information necessary to recover the cohomology ring H ∗(E, E#; R) from H ∗(X; R). Lemma. If π : E → X is a d-dimensional R-module vector bundle with d odd, then 2e(E) = 0 ∈ H d(X; R). Proof. Consider the map α : E → E given by negation on each fiber. This then gives an isomorphism a... |
∗(X) ⊗ H ∗(Rd, Rd \ {0}) ∼= H ∗(X × Rd, X × (Rd \ {0})) is an isomorphism. Then the claims of the Thom isomorphism theorem follow immediately. (i) For i < d, all the summands corresponding to H i(X × Rd, X × (Rd \ {0})) vanish since the H ∗(Rd, Rd \ {0}) term vanishes. (ii) The only non-vanishing summand for H d(X × R... |
uE|V ) ∈ H d(E|U, E#|U ) ⊕ H d(E|V, E#|V ) gets sent to 0 by i∗ V. Indeed, both the restriction of uE|U and uE|V to U ∩ V are Thom classes, so they are equal by uniqueness, so the difference vanishes. U − i∗ Then by exactness, there must be some uE|U ∪V ∈ H d(E|U ∪V, E#|U ∪V ) that restricts to uE|U and uE|V in U and V... |
0}, we know the inclusion j : S(E) E# is a homotopy equivalence, with inverse given by normalization. The long exact sequence for the pair (E, E#) gives (as before, we do not write the R): H i+d(E, E#) H i+d(E) H i+d(E#) H i+d+1(E, E#) Φ π∗ s∗ 0 j∗ Φ H i(X) · e(E) H i+d(X) p∗ H i+d(S(E)) p! H i+1(X) where p : S(E) → E ... |
isomorphisms H 0(CPn) e(L) H 2(CPn) e(L) H 4(CPn(L) Z · e(L)2 Similarly, we know that the terms in the odd degree vanish. Checking what happens at the end points carefully, the conclusion is that H ∗(CPn) = Z[e(L)]/(e(L)n+1) as a ring. Example. We do the real case of the above computation. We have K = γR 1,n+1 → RPn =... |
In this case, dϕ also has support in S. Note that this has a slight subtlety. The definition only requires that if σ lies completely outside S, then ϕ(σ) vanishes. However, we can have simplices that extends very far and only touches S slightly, and the support does not tell us anything about the value of σ. Later, we ... |
ian groups Gi for each i ∈ I and homomorphisms for all i, j ∈ I such that i ≤ j, such that ρij : Gi → Gj and ρii = idGi ρik = ρjk ◦ ρij whenever i ≤ j ≤ k. We define the direct limit on the system (Gi, ρij) to be Gi = lim −→ i∈I i∈I Gi /x − ρij(x) : x ∈ Gi. The underlying set of it is Gi /{x ∼ ρij(x) : x ∈ Gi}. i∈I In t... |
it. Lemma. We have H i c(Rd; R) ∼= R i = d 0 otherwise. Proof. Let B ∈ K(Rd) be the balls, namely B = {nDd, n = 0, 1, 2, · · · }. Then since every compact set is contained in one of them, we have H n c (X) ∼= lim −→ K∈K(Rd) H n(Rd, Rd \ K; R) ∼= lim −→ nDd∈B H n(Rd, Rd \ nDd; R) We can compute that directly. Since Rd ... |
e duality III Algebraic Topology Definition (Proper map). A map f : X → Y of spaces is proper if the preimage of a compact space is compact. Now if f is proper, then it does induce a map H ∗ c ( · ) by the usual construction. From now on, we will assume all spaces are Hausdorff, so that all compact subsets are closed. Th... |
-supported coho- mology. But before we do that, we first do it for local cohomology. Proposition. Let K, L ⊆ X be compact. Then there is a long exact sequence H n(X | K ∩ L) H n(X | K) ⊕ H n(X | L) H n(X | K ∪ L) H n+1(X | K ∩ L) ∂ H n+1(X | K) ⊕ H n+1(X | L) · · ·, where the unlabelled maps are those induced by inclusi... |
theorem, the right hand object gives the same (co)homology as C·(X \ K ∩ L; R). So we can produce another short exact sequence: 0 0 Hom C·(x) CU· (X\(K∩L)), R C·(X) Hom(C U· (X \ K ∩ L), R) C·(X, X \ K ∩ L) C·(X) Hom(C·(X \ K ∩ L), R) 0 0 Now the two right vertical arrows induce isomorphisms when we pass on to homolog... |
K) ⊕ H n+1(B | L) · · · ∂ The next step is to take the direct limit over K ∈ K(A) and L ∈ K(B). We need to make sure that these do indeed give the right compactly supported cohomology. The terms H n(A | K) ⊕ H n(B | L) are exactly right, and the one for H n(A ∩ B | K ∩ L) also works because every compact set in A ∩ B ... |
know that Hi(M | x; R) ∼= R i = d i = d 0. We can then define a local orientation of a manifold to be a generator of this group. Definition (Local R-orientation of manifold). Fr a d-manifold M, a local Rorientation of M at x is an R-module generator µx = Hd(M | x; R). Definition (R-orientation). An R-orientation of M is ... |
Uα ∩ Uβ) ⊆ Rd ϕ−1 is orientation-preserving, then M is R-orientable. 74 11 Manifolds and Poincar´e duality III Algebraic Topology Proof. (i) F2 has a unique F2-module generator. (ii) For x ∈ Uα, we define µx to be the image of the standard orientation of Rd via Hd(M | x) ∼= Hα(Ud | x) (ϕα)∗ Hd(Rd | ϕ−1 α (x)) Rd(Rd | 0)... |
by scaling away from x. Thus the map Hi(Rd | A) → Hi(Rd | x) is an isomorphism by the five lemma for all i. Then in degree d, there is some µA corresponding to µx. This µA is then has the required property by definition of orientability. The second part of the theorem also follows by what we know about Hi(Rd | x). Claim... |
we can find a good B such that A ⊆ B ⊆ Rd \ C. Then c ∈ Ci(Rd | B) is a cycle. So we know [c] ∈ Hi(Rd | B). So the map is surjective. Injectivity is obvious. An immediate consequence of this is that for i > d, we have Hi(Rd | A) = 0. Also, if i = d, we know that µA is given uniquely by the collection {µB}B∈G (uniquenes... |
> d. Moreover, if M itself is compact, then we know H 0 c (M ; R) has a special element 1. So we also get a canonical element of Hd(M ; R). But we know there is a special element of Hd(M ; R), namely the fundamental class. They are in fact the same, and this is no coincidence. This is in fact how we are going to produ... |
, the following diagram commutes: Hk(Y ; R) × H (Y ; R) Hk−(Y ; R) f∗×id Hk(X; R) × H (Y ; R) id ×f ∗ f∗ Hk(X; R) × H (X; R) Hk−(X; R) Proof. We check this on the cochain level. We let x = σ : ∆k → X. Then we have f#(σ f #y) = f# (f #y)(σ|[v0,...,v])σ|[v,...,vk] = y(f#(σ|[v0,...,v]))f#(σ|[v,...,vk]) = y((f#σ)|[v0,...,v... |
Manifolds and Poincar´e duality III Algebraic Topology So we get a duality map DM = lim −→ (µK · ) : lim −→ H (M | K; R) → Hd−(M ; R). Now we can prove Poincar´e duality. Proof. We say M is “good” if the Poincar´e duality theorem holds for M. We now do the most important step in the proof: Claim 0. Rd is good. The onl... |
· · · with M = good. n Un, and Ui are all good, then M is Any compact set in M lies in some Un, so the map lim −→ H c (Un) → H c (Un) is an isomorphism. Similarly, since simplices are compact, we also have Hd−k(M ) = lim −→ Hd−k(Un). Since the direct limit of open sets is open, we are done. Claim 3. Any open subset of... |
ϕ ∈ C k(X; R) and ψ ∈ C (X; R), then ψ(σ ϕ) = ψ(ϕ(σ|[v0,...,vk])σ|[vk,...,vk+]) = ϕ(σ|[v0,...,vk])ψ(σ|[vk,...,vk+]) = (ϕ ψ)(σ), Since we like diagrams, we can express this equality in terms of some diagram commuting. The map h : H k(X; R) → HomR(Hk(X; R), R) in the universal coefficients theorem is given by [ϕ] → ([σ] →... |
are isomorphisms. Proof. We have as we know ϕ, ψ = (−1)|ϕ||ψ|ψ, ϕ, ϕ ψ = (−1)|ϕ||ψ|ψ ϕ. So if one adjoint is an isomorphism, then so is the other. To see that they are isomorphsims, we notice that we have an isomorphism H k(M ; R) UCT HomR(Hk(M ; R), R) D∗ m HomR(H d−k(M ; R), R). [ϕ] ([σ] → ϕ(σ)) ([ψ] → ϕ([M ] ψ)) Bu... |
1 generates H 2n−2(CPn, Z), since it pulls back under i∗ to xn−1, which is a generator. Finally, consider H 2(CP2, Z) ⊗ H 2n−2(CP2, Z) Z y ⊗ yn−1 yn[CPn]. Since this is non-singular, we know yn ∈ H 2n(CPn, Z) must be a generator. Of course, we can get H ∗(RPn, F2) similarly. 11.4 Applications We go through two rather s... |
1-dimensional manifold W with boundary, then sgn(M ) = 0. Example. CP2 has H 2(CP2; R) ∼= R, and the bilinear form is represented by the matrix (1). So the signature is 1. So CP2 is not the boundary of a manifold. Degree Recall we defined the degree of a map from a sphere to itself. But if we have a Z-oriented space, w... |
[M ]) = (f ∗(α) f ∗(β))([M ]) = 0. This is a contradiction. 83 11 Manifolds and Poincar´e duality III Algebraic Topology 11.5 Intersection product Recall that cohomology comes with a cup product. Thus, Poincar´e duality gives us a product on homology. Our goal in this section is to understand this product. We restrict ... |
V # x ; R) ∼= H n(Ux, U # x ; R) ⊗ H m(Wx, W # x ; R) ∼= R. So any local R-orientation on any two induces one on the third, and it is straightforward to check the local compatibility condition. Can we find a more concrete description of this orientation on νN ⊆M? By c(Rd), we know the same argument as when we showed th... |
−n c (U ; R) extension by 0 H d−n(M ; R) The commutativity of the square is a straightforward naturality property of Poincar´e duality. Under the identification H d−n c (U ; R) ∼= H d−n(νN ⊆M, ν# N ⊆M ; R), the above (U ; R) is the Thom class of the says that the image of [N ] ∈ Hn(N ; R) in H d−n normal bundle νN ⊆M. c... |
�M ) ⊕ i∗ W (νW ⊆M ). So with some abuse of notation, we can write N EN ⊆M i∗ i∗ W EW ⊆M ∈ H ∗(νN ∩W ⊆M, ν# N ∩W ⊆M ; R), and we can check this gives the Thom class. So we have D−1 M ([N ]) D−1 M ([W ]) = D−1 M ([N ∩ W ]). The slogan is “cup product is Poincar´e dual to intersection”. One might be slightly worried abou... |
a fact that any two embeddings of CPk → CPn are homotopic, so we can choose these. Now these two manifolds intersect transversely, and the intersection is So this says that CPk ∩ CP = CPk+−n. yk · y = ±yl+−n, where there is some annoying sign we do not bother to figure out. So if xk is Poincar´e dual to yn−k, then xk x... |
do some magic that involves the diagonal map ∆ : M M × M x (x, x). This gives us a cohomology class δ = D−1 M ×M (∆∗[M ]) ∈ H d(M × M, Q) ∼= i+j=d H i(M, Q) ⊗ H j(M, Q). It turns out a lot of things we want to do with this δ can be helped a lot by doing the despicable thing called picking a basis. We let {ai} be a bas... |
k a)[M ] = (−1)|a||bk|δk. So we see that Ck = δk(−1)|a|. 88 11 Manifolds and Poincar´e duality III Algebraic Topology Corollary. We have the Euler characteristic. ∆∗(δ)[M ] = χ(M ), Proof. We note that for a ⊗ b ∈ H n(M × M ), we have ∆∗(a ⊗ b) = ∆∗(π∗ 1a π∗ 2b) = a b because πi ◦ ∆ = id. So we have ∆∗(δ) = (−1)|ai|ai ... |
e(E ⊕ F ) = e(E) e(F ). Proof. More precisely, we have projection maps πE E E ⊕ F 89 πF. F 11 Manifolds and Poincar´e duality III Algebraic Topology We let U = π−1 E (E#) and V = π−1 F (F #). Now observe that U ∪ V = (E ⊕ F )#. So if dim E = e, dim F = f, then we have a map H e(E, E#) ⊗ H f (F, F #) π∗ E ⊗π∗ F H e(E ⊕... |
condition is that the graph Γf = {(x, f (x)) ∈ M × M } ⊆ M × M has to be transverse to the diagonal. Since Γf ∩ ∆ is exactly the fixed points of f, this is equivalent to requiring that for each fixed point x, the map Dx∆ ⊕ DxF : TxM ⊕ TxM → TxM ⊕ TxM is an isomorphism. We can write the matrix of this map, which is I Dxf... |
Then we have (F ∗δ)[M ] = i,j (−1)|ai|Cij(ai ⊗ bj)[M ] = (−1)|ai|Cii, i and Cii is just the trace of f ∗. We now compute this product in a different way. As Γf and ∆(M ) are transverse, we know Γf ∩ ∆(M ) is a 0-manifold, and the orientation of Γf and ∆(M ) induces an orientation of it. So we have [Γf ] · [∆(m)] = [Γf ... |
, 6 Brouwer’s fixed point theorem, 20 cap product, 77 cell complex, 38 finite, 38 finite-dimensional, 38 subcomplex, 38 cellular cohomology, 41 cellular homology, 40 chain complex, 5 chain homotopy, 29 chain map, 6 characteristic class, 63 characteristic map, 38 coboundary, 6 cochain complex, 5 cochain maps, 6 cocycle, 6 ... |
zero, 56 short exact sequence, 15 signature of manifold, 83 singular n-simplex, 8 singular cohomology, 9 singular homology, 9 small simplices theorem, 32, 36 snake lemma, 25 sphere bundle, 66 standard n-simplex, 7 subcomplex, 38 support, 68 cochain, 68 tangent bundle, 58 Thom class, 62 transverse intersection, 85 tubu... |
DCC, we may choose n large enough so that f n : f n(M ) → f 2n(M ) is an isomorphism, as if we keep iterating f, the image is a descending chain and the kernel is an ascending chain, and these have to terminate. If m ∈ M, then we can write for some m1. Then f n(m) = f 2n(m1) m = f n(m1) + (m − f n(m1)) ∈ im f n + ker ... |
A. The projection onto Pj is achieved by left multiplication by an idempotent ej, Pj = ejA. The fact that the A decomposes as a direct sum of the Pj translates to the condition ej = 1, eiej = 0 for i = j. Definition (Orthogonal idempotent). A collection of idempotents {ei} is orthogonal if eiej = 0 for i = j. The indec... |
image fi−1 in A/N i−1. Then since x2 − x vansishes in A/N i−1, we know x2 − x ∈ N i−1/N i. Then in particular, (x2 − x)2 = 0 ∈ A/N i. (†) 24 1 Artinian algebras III Algebras We let fi = 3x2 − 2x3. Then by a direct computation using (†), we find f 2 i = fi, and fi has image 3fi−1 − 2fi−1 = fi−1 in A/N i−1 (alternatively... |
= f1 + · · · + ft ∈ A/J(A), and these fi are orthogonal primitive idempotents. Idempotent lifting then gives 1 = e1 + · · · + et ∈ A, and these are orthogonal primitive idempotents. So we can write A = ejA = Pi, 25 1 Artinian algebras III Algebras where Pi = eiA are indecomposable projectives, and Pi/PiJ(A) = Si is si... |
�. Conversely, suppose there is a non-zero map f : P → M. Then it factors as S = P P J(A) → im f (im f )J(A). Now we cannot have im f = (im f )J(A), or else we have im f = (im f )J(A)n = 0 for sufficiently large n since J(A) is nilpotent. So this map must be injective, hence an isomorphism. So this exhibits S as a compos... |
P1 ⊕ P2]. The abelian group is K0(A). Example. If A is an Artinian algebra, then we know that any finitely-generated projective is a direct sum of indecomposable projectives, and this decomposition is unique by Krull-Schmidt. So K0(A) = abelian group generated by the isomorphism classes of indecomposable projectives. So... |
trace of an endomorphism of P is well-defined in A/[A, A], independent of the choice of e. Thus we have a trace map EndA(P ) → A/[A, A]. In particular, the trace of the identity map on P is the trace of e. We call this the trace of P. Note that if we have finitely generated projectives P1 and P2, then we have P1 ⊕ Q1 = ... |
natural inclusion. This is the start of the theory of algebraic K-theory, which is a homology theory telling us about the endomorphisms of free A-modules. We can define K1(A) to be the abelianization of GL(A) = lim n→∞ GLn(A). K2(A) tells us something about the relations required if you express GL(A) in terms of genera... |
also Noetherian. The proof is very similar, but for completeness, we will spell it out completely. Theorem. Let A be left Noetherian. Then A[[X]] is Noetherian. Proof. Let I be a left ideal of A[[X]]. We’ll show that if A is left Noetherian, then I is finitely generated. Let Jr = {a : there exists an element of I of th... |
ideal generated by our list, and hence so is f. Example. It is straightforward to see that quotients of Noetherian algebras are Noetherian. Thus, algebra images of the algebras A[x] and A[[x]] would also be Noetherian. For example, finitely-generated commutative k-algebras are always Noetherian. Indeed, if we have a ge... |
, f ] = h, [h, e] = 2e, [h, f ] = −2f, To prove that An(k) and U(g) are Noetherian, we need some machinery, that involves some “deformation theory”. The main strategy is to make use of a natural filtration of the algebra. Definition (Filtered algebra). A (Z-)filtered algebra A is a collection of k-vector spaces · · · ≤ A−... |
understand a filtered algebra, we consider a nicer object, known as the associated graded algebra. Definition (Associated graded algebra). Given a filtration of A, the associated graded algebra is the vector space direct sum gr A = Ai Ai−1. This is given the structure of an algebra by defining multiplication by (a + Ai−1)... |
as for a finitely-generated algebra. Now observe that if ai ∈ Ai, and aj ∈ Aj, then So we see that gr A is commutative, and in fact aiaj − ajai ∈ Ai+j−2. gr An(k) ∼= k[ ¯X1, · · ·, ¯Xn, ¯Y1, · · ·, ¯Yn], where ¯Xi, ¯Yi are the images of Xi and Yi in A1/A0 respectively. This is not hard to prove, but is rather messy. It... |
1 < I2 < · · · of left ideals. Since we have a positive filtration, for some Ai, we have I1 ∩ Ai I2 ∩ Ai and I1 ∩ Ai−1 = I2 ∩ Ai−1. Thus gr I1 gr I2 gr I3 · · ·. This is a contradiction since gr A is Noetherian. So A must be Noetherian. Where we need positivity is the existence of that transition from equality to non-eq... |
= 1, XY = qY X. The word “quantum” in this context is usually thrown around a lot, and doesn’t really mean much apart from non-commutativity, and there is very little connection with actual physics. These algebras are both left and right Noetherian. We cannot prove these by filtering, as we just did. We will need a ver... |
(Poisson algebra). An associative algebra B is a Poisson algebra if there is a k-bilinear bracket { ·, · } : B × B → B such that – B is a Lie algebra under { ·, · }, i.e. {r, s} = −{s, r} and {{r, s}, t} + {{s, t}, r} + {{t, r}, s} = 0. – We have the Leibnitz rule {r, st} = s{r, t} + {r, s}t. The second condition says... |
ve got a finitely-generated graded module over a graded commutative algebra. To understand this further, we prove some results about graded modules over commutative algebras, which is going to apply to our gr A and gr M. Definition (Poincar´e series). Let V be a graded module over a graded algebra S, say ∞ V = Vi. Then t... |
Vi) + dim(Vi+km) − dim(Li+km ) = 0. We multiply by ti+km, and sum over i to get tkmP (K, t) − tkm P (V, t) + P (V, t) − P (L, t) = g(t), where g(t) is a polynomial with integral coefficients arising from consideration of the first few terms. We now apply the induction hypothesis to K and L, and we are done. Corollary. If ... |
omial in Q[t] of degree d. This φ(t) is the Hilbert polynomial, and χ(t) the Samuel polynomial. Some people call χ(t) the Hilbert polynomial instead, though. We now want to apply this to our cases of gr A, where A = An(k) or U(g), filtered as before. Then we deduce that i 0 dim Mj Mj−1 = χ(i), for a polynomial χ(t) ∈ Q[... |
. However, for general M = A, we can get non-integral values. In fact, the values we can get are 0, 1, 2, and then any real number ≥ 2. We can also have ∞ if the lim sup doesn’t exist. Example. If A = kG, then we have GK-dim(kG) < ∞ iff G has a subgroup H of finite index with H embedding into the strictly upper triangula... |
module matches the one we used when thinking about the polynomial algebra as a module over itself. So we get d(k[X1, · · ·, Xn]) = n. Lemma. Let M be a finitely-generated An-module. Then d(M ) ≤ 2n. Proof. Take generators m1, · · ·, ms of M. Then there is a surjective filtered module homomorphism An ⊕ · · · ⊕ An M (a1, ... |
So we get that n ≤ d(M ). So it remains to prove the claim. It suffices to prove that the natural map Ai → Homk(Mi, M2i), given by multiplication is injective. So we want to show that if a ∈ Ai = 0, then aMi = 0. We prove this by induction on i. When i = 0, then A0 = k, and M0 is a finite-dimensional k-vector space. Then... |
then we can form gr I and we can consider the set of maximal ideals containing it. This gives us the characteristic variety Ch(An/I). We saw that there was a Poisson bracket on gr An, and this may be used to define a skew-symmetric form on the tangent space at any point of the cotangent bundle. In this case, this is no... |
into a direct sum of matrix algebras over division algebras. But we actually cannot. We will have to first quotient End(E(A)) by some ideal I. On the other hand, we do not actually have an embedding of A ∼= EndA(A) into End(E(A)). Instead, what we have is only a homomorphism EndA(A) → End(E(A))/I, where we quotient out... |
Since M is injective as a k-module, we can find the β such that α = βθ. We define ψ : N1 → Homk(A, M ) by ψ(n1)(x) = β(n1x). It is straightforward to check that this does the trick. Also, we have an embedding M → Homk(A, M ) by m → (φn : x → mx). 43 2 Noetherian algebras III Algebras The category theorist will write the... |
a submodule of E. So (N ∩ E) ∩ M = N ∩ M = 0. So F is an essential extension of M. Lemma. A maximal essential extension is an injective module. Such maximal things exist by Zorn’s lemma. Proof. Let E be a maximal essential extension of M, and consider any embedding E → F. We shall show that E is a direct summand of F.... |
direct sums are the same as direct products), and also M1 ⊕ M2 embeds in E(M1) ⊕ E(M2). So it suffices to prove this extension is essential. Let V ≤ E(M1) ⊕ E(M2). Then either V /E(M1) = 0 or V /E(M2) = 0. We wlog it is the latter. Note that we can naturally view V E(M2) ≤ E(M1) ⊕ E(M2) E(M2) ∼= E(M1). Since M1 ⊆ E(M1) ... |
V ∩B = {0} since the extension is essential. So we have two non-zero submodules of V that intersect trivially. Conversely, suppose V is not uniform, and let V1, V2 be non-zero submodules that intersect trivially. By Zorn’s lemma, we suppose these are maximal submodules that intersect trivially. We claim E(V1) ⊕ E(V2) ... |
e. E(AA) is indecomposable. 46 2 Noetherian algebras III Algebras Proof. Suppose not, and so there are xA and yA non-zero such that xA ∩ yA = {0}. So xA ⊕ yA is a direct sum. But A is a domain and so yA ∼= A as a right A-module. Thus yxA ⊕ yyA is a direct sum living inside yA. Further decomposing yyA, we find that xA ⊕ ... |
, then ker(f ◦ g) ≥ ker g, and is hence also essential. So I is a maximal left ideal. The point of this lemma is to allow us to use Krull–Schmidt. Lemma. Let M be a non-zero Noetherian module. Then M is an essential extension of a direct sum of uniform submodules N1, · · ·, Nr. Thus E(M ) ∼= E(N1) ⊕ · · · E(Nr) is a di... |
is true for An(k) and U(g). Lemma. Let E1, · · ·, Er be indecomposable injectives. Put E = E1 ⊕ · · · ⊕ Er. Let I = {f ∈ EndA(E) : ker f is essential}. This is an ideal, and then EndA(E)/I ∼= Mn1(D1) ⊕ · · · ⊕ Mns (Ds) for some division algebras Di. Proof. We write the decomposition instead as E = En1 1 ⊕ · · · ⊕ Enr ... |
a finite direct sum of matrix algebras over division algebras. Proof. As usual, we have a map A x EndA(AA) left multiplication by x For a map AA → AA, it lifts to a map E(AA) → E(AA) by injectivity: 0 AA f AA θ θ E(AA) f E(AA) We can complete the diagram to give a map f : E(AA) → E(AA), which restricts to f on AA. This... |
, 1, 2, and we will see that these groups can be interpreted as things we are already familiar with. The construction of these Hochschild cohomology groups might seem a bit arbitrary. It is possible to justify these a priori using the general theory of homological algebra and/or model categories. On the other hand, Hoc... |
⊗k A is free on a single generator 1 ⊗k 1, whereas if {xi} is a k-basis of A, then A ⊗k A ⊗k A is free on {1 ⊗k xi ⊗k 1}. The general theory of homological algebra says we should be interested in such free things. Definition (Free resolution). Let A be an algebra and M an A-A-bimodule. A free resolution of M is an exac... |
A M and HomA-A( ·, M ) to any projective resolution of AAA and take the (co)homology, and the resulting vector spaces will be the same. However, we will not prove that, and will just always stick to this standard free resolution all the time. 3.2 Cohomology As mentioned, the construction of Hochschild cohomology involv... |
�� · · · ⊗ aiai+1 ⊗ · · · ⊗ an) + (−1)n+1f (a1 ⊗ · · · ⊗ an−1)an 51 3 Hochschild homology and cohomology III Algebras The reason the end ones look different is that we replaced HomA-A(A⊗(n+2), M ) with Homk(A⊗n, M ). The crucial observation is that the exactness of the Hochschild chain complex, and in particular the fac... |
resolution. However, since we don’t want to prove such general results, we shall provide an explicit computation. Proof. If AAA is projective, then all A⊗n are projective. At each degree n, we can split up the Hochschild chain complex as the short exact sequence 0 A⊗(n+3) ker dn dn A⊗(n+2) dn−1 im dn−1 0 The im d is a... |
of A ⊗ A. Since A ⊗ A is a free A-A-bimodule, this condition is equivalent to A being projective. However, there is some point in writing the definition like this. Note that an A-A-bimodule is equivalently a left A ⊗ Aop-module. Then AAA is a direct summand of A ⊗ A if and only if there is a separating idempotent e ∈ A... |
M ) d∗ −→ · · · Now d∗ sends Hom(ker µ, M ) to zero. So Hom(ker µ, M ) must be in the image of (∗). So the map Hom(A ⊗ A, M ) −→ Hom(ker µ, M ) must be surjective. This is true for any M. In particular, we can pick M = ker µ. Then the identity map idker µ lifts to a map A ⊗ A → ker µ whose restriction to ker µ is the ... |
for all a ∈ A}. In particular, HH 0(A, A) is the center of A. 54 3 Hochschild homology and cohomology III Algebras Proposition. ker δ1 = {f ∈ Homk(A, M ) : f (a1a2) = a1f (a2) + f (a1)a2}. These are the derivations from A to M. We write this as Der(A, M ). On the other hand, im δ0 = {f ∈ Homk(A, M ) : f (a) = am − ma ... |
Previously, we saw first cohomology can be understood in terms of derivations. We can formulate derivations in terms of this semi-direct product. Lemma. We have Derk(A, M ) ∼= {algebra complements to M in A M isomorphic to A}. 55 3 Hochschild homology and cohomology III Algebras Proof. A complement to M is an embedded ... |
be the canonical quotient map. Then we have a short exact sequence 0 I B A 0. Then two extensions B1 and B2 are isomorphic if there is a k-algebra isomorphism θ : B1 → B2 such that the following diagram commutes: 0 I B1 θ B2 A. 0 Note that the semi-direct product is such an extension, called the split extension. 56 3 ... |
-bimodule action in I. Thus, we find fρ − fρ = δ1(ρ − ρ), noting that ρ − ρ actually maps to I. So we obtain a map from the isomorphism classes of extensions to the second cohomology group. Conversely, given an A-A-bimodule M and a 2-cocycle f : A ⊗ A → M, we let as k-vector spaces. We define the multiplication map Bf = ... |
derburn, Malcev). Let B be a k-algebra satisfying – Dim(B/J(B)) ≤ 1. – J(B)2 = 0 Then there is an subalgebra A ∼= B/J(B) of B such that B = A J(B). Furthermore, if Dim(B/J(B)) = 0, then any two such subalgebras A, A are conjugate, i.e. there is some x ∈ J(B) such that Notice that 1 + x is a unit in B. A = (1 + x)A(1 + ... |
under direct sums, we know B/J(B) is k-separable, hence has dimension zero. It is a general fact that J(B) is nilpotent. 58 3 Hochschild homology and cohomology III Algebras 3.3 Star products We are now going to do study some deformation theory. Suppose A is a k-algebra. We write V for the underlying vector space of A... |
what elements of V = A are sent to. Let a, b ∈ V = A. We write f (a, b) = ab + tF1(a, b) + t2F2(a, b) + · · ·. Because of bilinearity, we know Fi are k-bilinear maps, and so correspond to k-linear maps V ⊗ V → V. For convenience, we will write F0(a, b) = ab. The only non-trivial requirement f has to satisfy is associa... |
0 Fm(Fn(a, b), c) − Fm(a, Fn(b, c)) = (δ2Fλ)(a, b, c). (†λ) Here we are identifying Fλ with the corresponding k-linear map A ⊗ A → A. For λ = 2, this says F1(F1(ab), c) − F1(a, F1(b, c)) = (δ2F2)(a, b, c). If F1 is a 2-cocycle, then one can check that the LHS gives a 3-cocycle. If F1 is integrable, then the LHS has to ... |
]]-linear automorphism Φ of V [[t]] of the form Φ(a) = a + tφ1(a) + t2φ2(a) + · · · sch that Equivalently, the following diagram has to commute: f (a, b) = Φ−1g(Φ(a), Φ(b)). V [[t]] ⊗ V [[t]] Φ⊗Φ V [[t]] ⊗ V [[t]] f g V [[t]] Φ V [[t]] Star products equivalent to the usual product on A ⊗ k[[t]] are called trivial. 60 3... |
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