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Because Ae is a direct sum of copies of C, Z is just a product of Riemann spheres, one for each ei. In particular it is compact. There is a natural map of Z into X which is continuous. | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
Let Y be the image of Z. It is compact and therefore coincides with the closure of Y0 = Ae β A = X0. The set Uβ
Y is the continuous image of the compact set U Γ Y. It is therefore compact. On the other hand, Uβ
Y0 = X0, so it contains a dense subset of X and must therefore coincide with X. So X is compact. | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
The above argument shows that every (a,b) in X is equivalent to k(c,d) with c and d in Ae and k in Ξu(A). The mapping of Z into X is in fact an embedding. This is a consequence of (x,y) being quasi-invertible in Ae if and only if it is quasi-invertible in A. Indeed, if B(x,y) is injective on A, its restriction to Ae is... | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
Let A be a finite-dimensional semisimple algebra, and { 0 } = J 0 β β― β J n β A {\displaystyle \{0\}=J_{0}\subset \cdots \subset J_{n}\subset A} be a composition series of A, then A is isomorphic to the following Cartesian product: A β J 1 Γ J 2 / J 1 Γ J 3 / J 2 Γ . . . | Wikipedia - Semisimple algebra | null | null | null |
Γ J n / J n β 1 Γ A / J n {\displaystyle A\simeq J_{1}\times J_{2}/J_{1}\times J_{3}/J_{2}\times ...\times J_{n}/J_{n-1}\times A/J_{n}} where each J i + 1 / J i {\displaystyle J_{i+1}/J_{i}\,} is a simple algebra. The proof can be sketched as follows. First, invoking the assumption that A is semisimple, one can show th... | Wikipedia - Semisimple algebra | null | null | null |
So J1 is a unital subalgebra and an ideal of J2. Therefore, one can decompose J 2 β J 1 Γ J 2 / J 1 . {\displaystyle J_{2}\simeq J_{1}\times J_{2}/J_{1}.} | Wikipedia - Semisimple algebra | null | null | null |
By maximality of J1 as an ideal in J2 and also the semisimplicity of A, the algebra J 2 / J 1 {\displaystyle J_{2}/J_{1}\,} is simple. Proceed by induction in similar fashion proves the claim. For example, J3 is the Cartesian product of simple algebras J 3 β J 2 Γ J 3 / J 2 β J 1 Γ J 2 / J 1 Γ J 3 / J 2 . | Wikipedia - Semisimple algebra | null | null | null |
{\displaystyle J_{3}\simeq J_{2}\times J_{3}/J_{2}\simeq J_{1}\times J_{2}/J_{1}\times J_{3}/J_{2}.} The above result can be restated in a different way. For a semisimple algebra A = A1 Γ...Γ An expressed in terms of its simple factors, consider the units ei β Ai. The elements Ei = (0,...,ei,...,0) are idempotent eleme... | Wikipedia - Semisimple algebra | null | null | null |
Let A be a finite-dimensional unital Jordan algebra over a field k of characteristic β 2. For a pair (a,b) with a and aβ1 β b invertible define In this case the Bergman operator B(a,b) = Q(a)Q(aβ1 β b) defines an invertible operator on A and In fact B ( a , b ) β 1 ( a β Q ( a ) b ) = Q ( a b ) Q ( a β 1 ) ( a β Q ( a ... | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
In that case Indeed, a b + c = ( ( a β 1 β b ) β c ) β 1 = ( ( a b ) β 1 β c ) β 1 = ( a b ) c . {\displaystyle \displaystyle {a^{b+c}=((a^{-1}-b)-c)^{-1}=((a^{b})^{-1}-c)^{-1}=(a^{b})^{c}.}} The assumption that a be invertible can be dropped since ab can be defined only supposing that the Bergman operator B(a,b) is in... | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
The pair (a,b) is then said to be quasi-invertible. In that case ab is defined by the formula a b = B ( a , b ) β 1 ( a β Q ( a ) b ) . {\displaystyle \displaystyle {a^{b}=B(a,b)^{-1}(a-Q(a)b).}} | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
If B(a,b) is invertible, then B(a,b)c = 1 for some c. The fundamental identity implies that B(a,b)Q(c)B(b,a) = I. So by finite-dimensionality B(b,a) is invertible. Thus (a,b) is invertible if and only if (b,a) is invertible and in this case In fact B(a,b)(a + Q(a)ba) = a β 2R(a,b)a + Q(a)Q(b)a + Q(a)(b β Q(b)a) = a β Q... | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
{\displaystyle \displaystyle {a^{b+c}=(a^{b})^{c}.}} If k = R or C, this would follow by continuity from the special case where a and aβ1 β b were invertible. In general the proof requires four identities for the Bergman operator: In fact applying Q to the identity B(a,b)ab = a β Q(a)b yields B ( a , b ) Q ( a b ) B ( ... | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
{\displaystyle \displaystyle {B(a,b)Q(a^{b})B(b,a)=B(a,b)Q(a)=Q(a)B(b,a).}} The first identity follows by cancelling B(a,b) and B(b,a). The second identity follows by similar cancellation in B(a,b)Q(ab,c)B(b,a) = Q(B(a,b)ab,B(a,b)c) = Q(a β Q(a)b,B(a,b)c) = B(a,b)(Q(a,c) β R(c,b)Q(a)) = (Q(a,c) β Q(a)R(b,c))B(b,a).The ... | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
Since this mutation might not necessarily unital this means that when an identity is adjoint 1 β a becomes invertible in Ab β k1. This condition can be expressed as follows without mentioning the mutation or homotope: In fact if (a,b) is quasi-invertible, then c = ab satisfies the first identity by definition. The seco... | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
Let A be a finite-dimensional unital Jordan algebra over a field k of characteristic β 2. Two pairs (ai,bi) with ai invertible are said to be equivalent if (a1)β1 β b1 + b2 is invertible and a2 = (a1)b1 β b2. This is an equivalence relation, since if a is invertible a0 = a so that a pair (a,b) is equivalent to itself. ... | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
It is transitive. For suppose that (a3,b3) is a third pair with (a2)β1 β b2 + b3 invertible and a3 = (a2)b2 β b3. | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
From the above a 1 β 1 β b 1 + b 3 = ( a 1 β 1 β b 1 + b 2 ) β b 2 + b 3 = a 2 β 1 β b 2 + b 3 {\displaystyle \displaystyle {a_{1}^{-1}-b_{1}+b_{3}=(a_{1}^{-1}-b_{1}+b_{2})-b_{2}+b_{3}=a_{2}^{-1}-b_{2}+b_{3}}} is invertible and a 3 = a 2 b 2 β b 3 = ( a 1 b 1 β b 2 ) b 2 β b 3 = a 1 b 1 β b 3 . {\displaystyle \displays... | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
Two pairs (ai,bi) are said to be equivalent if (a1, b1 β b2) is quasi-invertible and a2 = (a1)b1 β b2. When k = R or C, the fact that this more general definition also gives an equivalence relation can deduced from the invertible case by continuity. For general k, it can also be verified directly: The relation is refle... | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
The relation is transitive. For suppose that (a3,b3) is a third pair with (a2, b2 β b3) quasi-invertible and a3 = (a2)b2 β b3. In this case B ( a 1 , b 1 β b 3 ) = B ( a 1 , b 1 β b 2 ) B ( a 2 , b 2 β b 3 ) , {\displaystyle \displaystyle {B(a_{1},b_{1}-b_{3})=B(a_{1},b_{1}-b_{2})B(a_{2},b_{2}-b_{3}),}} so that (a1,b1 ... | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
Let A be a graded algebra over a field k. If V is a finite-dimensional generating subspace of A, then we let f ( n ) = dim k β‘ V n {\displaystyle f(n)=\dim _{k}V^{n}} and then put It is called the GelfandβKirillov dimension of A. It is easy to show gk β‘ ( A ) {\displaystyle \operatorname {gk} (A)} is independent of a c... | Wikipedia - Dimension theory (algebra) | null | null | null |
Let A be a not-necessarily-commutative algebra over a field k. Even if A is not commutative, it can still happen that A has a Z {\displaystyle \mathbb {Z} } -filtration so that the associated ring gr β‘ A = β¨ i = β β β A i / A i β 1 {\displaystyle \operatorname {gr} A=\bigoplus _{i=-\infty }^{\infty }A_{i}/{A_{i-1}}} is... | Wikipedia - Algebraic set | null | null | null |
Let A be a propositional formula. The GΓΆdelβTarski translation of A is defined recursively as follows: T ( p n ) = β» p n {\displaystyle T(p_{n})=\Box p_{n}} T ( Β¬ A ) = β» Β¬ T ( A ) {\displaystyle T(\neg A)=\Box \neg T(A)} T ( A β§ B ) = T ( A ) β§ T ( B ) {\displaystyle T(A\land B)=T(A)\land T(B)} T ( A β¨ B ) = T ( A ) β¨... | Wikipedia - GΓΆdelβDummett logic | null | null | null |
Let A be a quadratic Jordan algebra over R or C. Following Jacobson (1969), a linear Jordan algebra structure can be associated with A such that, if L(a) is Jordan multiplication, then the quadratic structure is given by Q(a) = 2L(a)2 β L(a2). Firstly the axiom Q(a)R(b,a) = R (a,b)Q(a) can be strengthened to Q ( a ) R ... | Wikipedia - Quadratic Jordan algebra | null | null | null |
{\displaystyle \displaystyle {2Q(a)Q(b,c)a=2Q(Q(a)c,a)b.}} Switching b and c then gives Q ( a ) R ( b , a ) c = 2 Q ( Q ( a ) b , a ) c . {\displaystyle \displaystyle {Q(a)R(b,a)c=2Q(Q(a)b,a)c.}} | Wikipedia - Quadratic Jordan algebra | null | null | null |
Now let L ( a ) = 1 2 R ( a , 1 ) . {\displaystyle \displaystyle {L(a)={\frac {1}{2}}R(a,1).}} | Wikipedia - Quadratic Jordan algebra | null | null | null |
Replacing b by a and a by 1 in the identity above gives R ( a , 1 ) = R ( 1 , a ) = 2 Q ( a , 1 ) . {\displaystyle \displaystyle {R(a,1)=R(1,a)=2Q(a,1).}} In particular L ( a ) = Q ( a , 1 ) , L ( 1 ) = Q ( 1 , 1 ) = I . | Wikipedia - Quadratic Jordan algebra | null | null | null |
{\displaystyle \displaystyle {L(a)=Q(a,1),\,\,\,L(1)=Q(1,1)=I.}} If furthermore a is invertible then R ( a , b ) = 2 Q ( Q ( a ) b , a ) Q ( a ) β 1 = 2 Q ( a ) Q ( b , a β 1 ) . {\displaystyle \displaystyle {R(a,b)=2Q(Q(a)b,a)Q(a)^{-1}=2Q(a)Q(b,a^{-1}).}} | Wikipedia - Quadratic Jordan algebra | null | null | null |
Similarly if 'b is invertible R ( a , b ) = 2 Q ( a , b β 1 ) Q ( b ) . {\displaystyle \displaystyle {R(a,b)=2Q(a,b^{-1})Q(b).}} The Jordan product is given by a β b = L ( a ) b = 1 2 R ( a , 1 ) b = Q ( a , b ) 1 , {\displaystyle \displaystyle {a\circ b=L(a)b={\frac {1}{2}}R(a,1)b=Q(a,b)1,}} so that a β b = b β a . | Wikipedia - Quadratic Jordan algebra | null | null | null |
{\displaystyle \displaystyle {a\circ b=b\circ a.}} The formula above shows that 1 is an identity. Defining a2 by aβa = Q(a)1, the only remaining condition to be verified is the Jordan identity = 0. | Wikipedia - Quadratic Jordan algebra | null | null | null |
{\displaystyle \displaystyle {=0.}} In the fundamental identity Q ( Q ( a ) b ) = Q ( a ) Q ( b ) Q ( a ) , {\displaystyle \displaystyle {Q(Q(a)b)=Q(a)Q(b)Q(a),}} Replace a by a + t, set b = 1 and compare the coefficients of t2 on both sides: Q ( a ) = 2 Q ( a , 1 ) 2 β Q ( a 2 , 1 ) = 2 L ( a ) 2 β L ( a 2 ) . {\displ... | Wikipedia - Quadratic Jordan algebra | null | null | null |
Let A be a quadratic Jordan algebra over a field k of characteristic β 2. Following Jacobson (1969), a linear Jordan algebra structure can be associated with A such that, if L(a) is Jordan multiplication, then the quadratic structure is given by Q(a) = 2L(a)2 β L(a2). Firstly the axiom Q(a)R(b,a) = R(a,b)Q(a) can be st... | Wikipedia - Isotope (Jordan algebra) | null | null | null |
Indeed, applied to c, the first two terms give 2 Q ( a ) Q ( b , c ) a = 2 Q ( Q ( a ) c , a ) b . {\displaystyle \displaystyle {2Q(a)Q(b,c)a=2Q(Q(a)c,a)b.}} Switching b and c then gives Q ( a ) R ( b , a ) c = 2 Q ( Q ( a ) b , a ) c . | Wikipedia - Isotope (Jordan algebra) | null | null | null |
{\displaystyle \displaystyle {Q(a)R(b,a)c=2Q(Q(a)b,a)c.}} Now let L ( a ) = 1 2 R ( a , 1 ) . {\displaystyle \displaystyle {L(a)={\frac {1}{2}}R(a,1).}} | Wikipedia - Isotope (Jordan algebra) | null | null | null |
{\displaystyle \displaystyle {L(a)=Q(a,1),\,\,\,L(1)=Q(1,1)=I.}} The Jordan product is given by a β b = L ( a ) b = 1 2 R ( a , 1 ) b = Q ( a , b ) 1 , {\displaystyle \displaystyle {a\circ b=L(a)b={\frac {1}{2}}R(a,1)b=Q(a,b)1,}} so that a β b = b β a . {\displaystyle \displaystyle {a\circ b=b\circ a.}} | Wikipedia - Isotope (Jordan algebra) | null | null | null |
The formula above shows that 1 is an identity. Defining a2 by aβa = Q(a)1, the only remaining condition to be verified is the Jordan identity = 0. {\displaystyle \displaystyle {=0.}} | Wikipedia - Isotope (Jordan algebra) | null | null | null |
In the fundamental identity Q ( Q ( a ) b ) = Q ( a ) Q ( b ) Q ( a ) , {\displaystyle \displaystyle {Q(Q(a)b)=Q(a)Q(b)Q(a),}} Replace a by a + t1, set b = 1 and compare the coefficients of t2 on both sides: Q ( a ) = 2 Q ( a , 1 ) 2 β Q ( a 2 , 1 ) = 2 L ( a ) 2 β L ( a 2 ) . {\displaystyle \displaystyle {Q(a)=2Q(a,1)... | Wikipedia - Isotope (Jordan algebra) | null | null | null |
Let A be a superalgebra over a commutative ring K. The submodule A0, consisting of all even elements, is closed under multiplication and contains the identity of A and therefore forms a subalgebra of A, naturally called the even subalgebra. It forms an ordinary algebra over K. The set of all odd elements A1 is an A0-bi... | Wikipedia - Even subalgebra | null | null | null |
Let A be a unital Jordan algebra over a field K of characteristic not equal to 2. For a in A, let L denote the left multiplication map in the associative enveloping algebra L ( a ): x β¦ a x {\displaystyle L(a):x\mapsto ax\ } and define a K-endomorphism of A, called the quadratic representation, by Q ( a ) = 2 L ( a ) 2... | Wikipedia - Quadratic Jordan algebra | null | null | null |
Let A be a unital Jordan algebra over a field k of characteristic β 2. An element a in a unital Jordan algebra A is said to be invertible if there is an element b such that ab = 1 and a2b = a.Properties. If ab = 1 and a2b = a, then Q(a)b = 2a(ab) β (a2)b = a. The Jordan identity = 0 can be polarized by replacing x by ... | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
{\displaystyle \displaystyle {+2=0.}} Taking x = a or b and y = b or a shows that L(a2) commutes with L(b) and L(b2) commutes with L(a). Hence (b2)(a2) = 1. | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
Applying L(b) gives b2a = b. Hence Q(a)b2 = 1. Conversely if Q(a)b = a and Q(a)b2 = 1, then the second relation gives Q(a)Q(b)2 Q(a) = I. So both Q(a) and Q(b) are invertible. The first gives Q(a)Q(b)Q(a) = Q(a) so that Q(a) and Q(b) are each other's inverses. | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
Since L(b) commutes with Q(b) it commutes with its inverse Q(a). Similarly L(a) commutes with Q(b). So (a2)b = L(b)a2 = Q(a)b = a and ab = L(b)Q(a)b= Q(a)Q(b)1= 1. | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
Indeed, if a is invertible then the above implies Q(a) is invertible with inverse Q(b). Any inverse b satisfies Q(a)b = a, so b = Q(a)β1a. Conversely if Q(a) is invertible let b = Q(a)β1a. | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
Then Q(a)b = a. The fundamental identity then implies that Q(b) and Q(a) are each other's inverses so that Q(a)b2 = Q(a)Q(b)1=1. This follows from the formula aβ1 = Q(a)β1a. Suppose that Q(a)c = 1. | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
Then by the fundamental identity Q(a) is invertible, so a is invertible. This is an immediate consequence of the fundamental identity and the fact that STS is invertible if and only S and T are invertible. In the commutation identity Q(a)R(b,a) = Q(Q(a)b,a), set b = c2 with c = aβ1. | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
Then Q(a)b = 1 and Q(1,a) = L(a). Since L(a) commutes with L(c2), R(b,a) = L(c) = L(aβ1). If L(a) and L(b) commute, then ba = 1 implies b(a2) = a. Conversely suppose that a is invertible with inverse b. Then ab = 1. | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
Morevoer L(b) commutes with Q(b) and hence its inverse Q(a). So it commutes with L(a) = Q(a)L(b). The algebra k is commutative and associative, so if b is an inverse there ab =1 and a2b = a. Conversely Q(a) leaves k invariant. So if it is bijective on A it is bijective there. Thus aβ1 = Q(a)β1a lies in k. | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
Let A be a unital Jordan algebra over a field k of characteristic β 2. For a in A define the Jordan multiplication operator on A by L ( a ) b = a b {\displaystyle \displaystyle {L(a)b=ab}} and the quadratic representation Q(a) by Q ( a ) = 2 L ( a ) 2 β L ( a 2 ) . {\displaystyle Q(a)=2L(a)^{2}-L(a^{2}).\,} It satisfie... | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
{\displaystyle R(a,b)c=2Q(a,c)b,\,\,\,Q(x,y)={\frac {1}{2}}(Q(x+y)-Q(x)-Q(y)).} In particular if a or b is invertible then R ( a , b ) = 2 Q ( a ) Q ( a β 1 , b ) = 2 Q ( a , b β 1 ) Q ( b ) . {\displaystyle \displaystyle {R(a,b)=2Q(a)Q(a^{-1},b)=2Q(a,b^{-1})Q(b).}} | Wikipedia - Hua's identity (Jordan algebra) | null | null | null |
Let A be a unital Jordan algebra. If a, b and a β b are invertible, then Hua's identity holds: In particular if x and 1 β x are invertible, then so too is 1 β xβ1 with ( 1 β x ) β 1 + ( 1 β x β 1 ) β 1 = 1. {\displaystyle \displaystyle {(1-x)^{-1}+(1-x^{-1})^{-1}=1.}} | Wikipedia - Mutation (Jordan algebra) | null | null | null |
To prove the identity for x, set y = (1 β x)β1. Then L(y) = Q(1 β x)β1L(1 β x). Thus L(y) commutes with L(x) and Q(x). | Wikipedia - Mutation (Jordan algebra) | null | null | null |
Since Q(y) = Q(1 β x)β1, it also commutes with L(x) and Q(x). Since L(xβ1) = Q(x)β1L(x), L(y) also commutes with L(xβ1) and Q(xβ1). It follows that (xβ1 β 1)xy =(1 β x) y = 1. | Wikipedia - Mutation (Jordan algebra) | null | null | null |
Moreover, y β 1 = xy since (1 β x)y = 1. So L(xy) commutes with L(x) and hence L(xβ1 β 1). Thus 1 β xβ1 has inverse 1 β y. Now let Aa be the mutation of A defined by a. The identity element of Aa is aβ1. | Wikipedia - Mutation (Jordan algebra) | null | null | null |
Moreover, an invertible element c in A is also invertible in Aa with inverse Q(a)β1 cβ1. Let x = Q(a)β1b in Aa. It is invertible in A, as is aβ1 β Q(a)β1b = Q(a)β1(a β b). So by the special case of Hua's identity for x in Aa a β 1 = Q ( a ) β 1 ( a β 1 β Q ( a ) β 1 b ) β 1 + Q ( a ) β 1 ( a β 1 β b β 1 ) β 1 = ( a β b... | Wikipedia - Mutation (Jordan algebra) | null | null | null |
Let A be a vector space over K with a quadratic form q and associated symmetric bilinear form q(x,y) = q(x+y) - q(x) - q(y). Let e be a "basepoint" of A, that is, an element with q(e) = 1. Define a linear functional T(y) = q(y,e) and a "reflection" yβ = T(y)e - y. For each x we define Q(x) by Q(x): y β¦ q(x,yβ)x β q(x) ... | Wikipedia - Quadratic Jordan algebra | null | null | null |
Let A be an algebra over a commutative ring R. Then the algebra A is a right module over A e := A o p β R A {\displaystyle A^{e}:=A^{op}\otimes _{R}A} with the action x β
( a β b ) = a x b {\displaystyle x\cdot (a\otimes b)=axb} . Then, by definition, A is said to separable if the multiplication map A β R A β A , x β y... | Wikipedia - Commutative algebra (structure) | null | null | null |
Let A be an algebra over a field F with multiplication (not assumed to be associative) denoted by juxtaposition. For an element a of A, define the left a-homotope A ( a ) {\displaystyle A(a)} to be the algebra with multiplication x β y = ( x a ) y . {\displaystyle x*y=(xa)y.\,} Similarly define the left (a,b) mutation ... | Wikipedia - Mutation (algebra) | null | null | null |
Let A be an arrival flow, arriving at the ingress of a server, and D be the flow departing at the egress. A backlog period is an interval I such that, on any t β I, A(t)>D(t). The system is said to provide a strict minimal service curve S to the pair (A,B) iff, β s , t β R + {\displaystyle \forall s,t\in \mathbb {R} ^{... | Wikipedia - Network calculus | null | null | null |
Let A be an arrival flow, arriving at the ingress of a server, and D be the flow departing at the egress. The system is said to provide a simple minimal service curve S to the pair (A,B), if for all t it holds that D ( t ) β₯ ( A β S ) ( t ) . {\displaystyle D(t)\geq (A\otimes S)(t).} | Wikipedia - Network calculus | null | null | null |
Let A be an associative algebra over a commutative ring R. Since A is in particular a module, we can take the dual module A* of A. A priori, the dual A* need not have a structure of an associative algebra. However, A may come with an extra structure (namely, that of a Hopf algebra) so that the dual is also an associati... | Wikipedia - Enveloping algebra of an associative algebra | null | null | null |
The "co-" refers to the fact that they satisfy the dual of the usual multiplication and unit in the algebra axiom. Hence, the dual A β {\displaystyle A^{*}} is an associative algebra. The co-multiplication and co-unit are also important in order to form a tensor product of representations of associative algebras (see Β§... | Wikipedia - Enveloping algebra of an associative algebra | null | null | null |
Let A be an associative algebra over the complex numbers. Then A is said to be quasi-free if the following equivalent conditions are met: Given a square-zero extension R β R / I {\displaystyle R\to R/I} , each homomorphism A β R / I {\displaystyle A\to R/I} lifts to A β R {\displaystyle A\to R} . The cohomological dime... | Wikipedia - Quasi-free algebra | null | null | null |
Let A be an m Γ n matrix and k an integer with 0 < k β€ m, and k β€ n. A k Γ k minor of A, also called minor determinant of order k of A or, if m = n, (nβk)th minor determinant of A (the word "determinant" is often omitted, and the word "degree" is sometimes used instead of "order") is the determinant of a k Γ k matrix o... | Wikipedia - Matrix of cofactors | null | null | null |
Let A be an ordered algebra with unit e and let C* denote the cone in A* (the algebraic dual of A) of all positive linear forms on A. If f is a linear form on A such that f(e) = 1 and f generates an extreme ray of C* then f is a multiplicative homomorphism. | Wikipedia - Ordered algebra | null | null | null |
Let A be the adjacency matrix of a graph. Then the graph is regular if and only if j = ( 1 , β¦ , 1 ) {\displaystyle {\textbf {j}}=(1,\dots ,1)} is an eigenvector of A. Its eigenvalue will be the constant degree of the graph. Eigenvectors corresponding to other eigenvalues are orthogonal to j {\displaystyle {\textbf {j}... | Wikipedia - K-regular graph | null | null | null |
The "only if" direction is a consequence of the PerronβFrobenius theorem.There is also a criterion for regular and connected graphs: a graph is connected and regular if and only if the matrix of ones J, with J i j = 1 {\displaystyle J_{ij}=1} , is in the adjacency algebra of the graph (meaning it is a linear combinatio... | Wikipedia - K-regular graph | null | null | null |
Let A i {\displaystyle {\mathfrak {A}}_{i}} be the von Neumann algebra of bounded operators on H i {\displaystyle H_{i}} for i = 1 , 2. {\displaystyle i=1,2.} Then the von Neumann tensor product of the von Neumann algebras is the strong completion of the set of all finite linear combinations of simple tensor products A... | Wikipedia - Tensor product of Hilbert spaces | null | null | null |
This is exactly equal to the von Neumann algebra of bounded operators of H 1 β H 2 . {\displaystyle H_{1}\otimes H_{2}.} Unlike for Hilbert spaces, one may take infinite tensor products of von Neumann algebras, and for that matter C*-algebras of operators, without defining reference states. This is one advantage of the... | Wikipedia - Tensor product of Hilbert spaces | null | null | null |
Let A x = b {\displaystyle Ax=b} be a system of linear equations, let m {\displaystyle m} be the number of rows of A, a i {\displaystyle a_{i}} be the i {\displaystyle i} th row of complex-valued matrix A {\displaystyle A} , and let x 0 {\displaystyle x^{0}} be arbitrary complex-valued initial approximation to the solu... | Wikipedia - Kaczmarz method | null | null | null |
Let A {\displaystyle A} , B {\displaystyle B} be two square matrices over a ring R {\displaystyle {\mathcal {R}}} , for example matrices whose entries are integers or the real numbers. The goal of matrix multiplication is to calculate the matrix product C = A B {\displaystyle C=AB} . The following exposition of the alg... | Wikipedia - Strassen algorithm | null | null | null |
The naive algorithm would be: = . {\displaystyle {\begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}}={\begin{bmatrix}A_{11}B_{11}+A_{12}B_{21}&A_{11}B_{12}+A_{12}B_{22}\\A_{21}B_{11}+A_{22}B_{21}&A_{21}B_{12}+A_{22}B_{22}\end{bmatrix}}.} This construction does not reduce the number of multiplications: 8 multip... | Wikipedia - Strassen algorithm | null | null | null |
We recursively iterate this division process until the submatrices degenerate into numbers (elements of the ring R {\displaystyle {\mathcal {R}}} ). If, as mentioned above, the original matrix had a size that was not a power of 2, then the resulting product will have zero rows and columns just like A {\displaystyle A} ... | Wikipedia - Strassen algorithm | null | null | null |
The particular crossover point for which Strassen's algorithm is more efficient depends on the specific implementation and hardware. Earlier authors had estimated that Strassen's algorithm is faster for matrices with widths from 32 to 128 for optimized implementations. However, it has been observed that this crossover ... | Wikipedia - Strassen algorithm | null | null | null |
Let A {\displaystyle A} be a commutative Banach algebra, defined over the field C {\displaystyle \mathbb {C} } of complex numbers. A non-zero algebra homomorphism (a multiplicative linear functional) Ξ¦: A β C {\displaystyle \Phi \colon A\to \mathbb {C} } is called a character of A {\displaystyle A} ; the set of all cha... | Wikipedia - Gelfand spectrum | null | null | null |
The space Ξ¦ A {\displaystyle \Phi _{A}} is compact (in the topology just defined) if and only if the algebra A {\displaystyle A} has an identity element. Given a β A {\displaystyle a\in A} , one defines the function a ^: Ξ¦ A β C {\displaystyle {\widehat {a}}:\Phi _{A}\to {\mathbb {C} }} by a ^ ( Ο ) = Ο ( a ) {\display... | Wikipedia - Gelfand spectrum | null | null | null |
This homomorphism is the Gelfand representation of A {\displaystyle A} , and a ^ {\displaystyle {\widehat {a}}} is the Gelfand transform of the element a {\displaystyle a} . In general, the representation is neither injective nor surjective. In the case where A {\displaystyle A} has an identity element, there is a bije... | Wikipedia - Gelfand spectrum | null | null | null |
Let A {\displaystyle A} be a commutative ring and M {\displaystyle M} an A-module. There are different equivalent definitions of a connection on M {\displaystyle M} . | Wikipedia - Connection (algebraic framework) | null | null | null |
Let A {\displaystyle A} be a finite-dimensional algebra over K . {\displaystyle K.} In particular, A {\displaystyle A} is a finite-dimensional vector-space over K . {\displaystyle K.} | Wikipedia - Ring of finite adeles | null | null | null |
As a consequence, A A {\displaystyle \mathbb {A} _{A}} is defined and A A β
A K β K A . {\displaystyle \mathbb {A} _{A}\cong \mathbb {A} _{K}\otimes _{K}A.} Since there is multiplication on A K {\displaystyle \mathbb {A} _{K}} and A , {\displaystyle A,} a multiplication on A A {\displaystyle \mathbb {A} _{A}} can be de... | Wikipedia - Ring of finite adeles | null | null | null |
{\displaystyle \forall \alpha ,\beta \in \mathbb {A} _{K}{\text{ and }}\forall a,b\in A:\qquad (\alpha \otimes _{K}a)\cdot (\beta \otimes _{K}b):=(\alpha \beta )\otimes _{K}(ab).} As a consequence, A A {\displaystyle \mathbb {A} _{A}} is an algebra with a unit over A K . {\displaystyle \mathbb {A} _{K}.} | Wikipedia - Ring of finite adeles | null | null | null |
Let B {\displaystyle {\mathcal {B}}} be a finite subset of A , {\displaystyle A,} containing a basis for A {\displaystyle A} over K . {\displaystyle K.} For any finite place v {\displaystyle v} , M v {\displaystyle M_{v}} is defined as the O v {\displaystyle O_{v}} -module generated by B {\displaystyle {\mathcal {B}}} ... | Wikipedia - Ring of finite adeles | null | null | null |
{\displaystyle A_{v}.} For each finite set of places, P β P β , {\displaystyle P\supset P_{\infty },} define A A ( P , Ξ± ) = β v β P A v Γ β v β P M v . {\displaystyle \mathbb {A} _{A}(P,\alpha )=\prod _{v\in P}A_{v}\times \prod _{v\notin P}M_{v}.} | Wikipedia - Ring of finite adeles | null | null | null |
One can show there is a finite set P 0 , {\displaystyle P_{0},} so that A A ( P , Ξ± ) {\displaystyle \mathbb {A} _{A}(P,\alpha )} is an open subring of A A , {\displaystyle \mathbb {A} _{A},} if P β P 0 . {\displaystyle P\supset P_{0}.} Furthermore A A {\displaystyle \mathbb {A} _{A}} is the union of all these subrings... | Wikipedia - Ring of finite adeles | null | null | null |
Let A {\displaystyle A} be a finite-dimensional algebra over K . {\displaystyle K.} Since A A Γ {\displaystyle \mathbb {A} _{A}^{\times }} is not a topological group with the subset-topology in general, equip A A Γ {\displaystyle \mathbb {A} _{A}^{\times }} with the topology similar to I K {\displaystyle I_{K}} above a... | Wikipedia - Ring of finite adeles | null | null | null |
The elements of the idele group are called idele of A . {\displaystyle A.} Proposition. | Wikipedia - Ring of finite adeles | null | null | null |
Let Ξ± {\displaystyle \alpha } be a finite subset of A , {\displaystyle A,} containing a basis of A {\displaystyle A} over K . {\displaystyle K.} For each finite place v {\displaystyle v} of K , {\displaystyle K,} let Ξ± v {\displaystyle \alpha _{v}} be the O v {\displaystyle O_{v}} -module generated by Ξ± {\displaystyle ... | Wikipedia - Ring of finite adeles | null | null | null |
{\displaystyle A_{v}.} There exists a finite set of places P 0 {\displaystyle P_{0}} containing P β {\displaystyle P_{\infty }} such that for all v β P 0 , {\displaystyle v\notin P_{0},} Ξ± v {\displaystyle \alpha _{v}} is a compact subring of A v . {\displaystyle A_{v}.} | Wikipedia - Ring of finite adeles | null | null | null |
Furthermore, Ξ± v {\displaystyle \alpha _{v}} contains A v Γ . {\displaystyle A_{v}^{\times }.} For each v , A v Γ {\displaystyle v,A_{v}^{\times }} is an open subset of A v {\displaystyle A_{v}} and the map x β¦ x β 1 {\displaystyle x\mapsto x^{-1}} is continuous on A v Γ . | Wikipedia - Ring of finite adeles | null | null | null |
{\displaystyle A_{v}^{\times }.} As a consequence x β¦ ( x , x β 1 ) {\displaystyle x\mapsto (x,x^{-1})} maps A v Γ {\displaystyle A_{v}^{\times }} homeomorphically on its image in A v Γ A v . {\displaystyle A_{v}\times A_{v}.} | Wikipedia - Ring of finite adeles | null | null | null |
For each v β P 0 , {\displaystyle v\notin P_{0},} the Ξ± v Γ {\displaystyle \alpha _{v}^{\times }} are the elements of A v Γ , {\displaystyle A_{v}^{\times },} mapping in Ξ± v Γ Ξ± v {\displaystyle \alpha _{v}\times \alpha _{v}} with the function above. Therefore, Ξ± v Γ {\displaystyle \alpha _{v}^{\times }} is an open and... | Wikipedia - Ring of finite adeles | null | null | null |
Let A {\displaystyle A} be a unital associative algebra. The parameter-independent YangβBaxter equation is an equation for R {\displaystyle R} , an invertible element of the tensor product A β A {\displaystyle A\otimes A} . The YangβBaxter equation is R 12 R 13 R 23 = R 23 R 13 R 12 , {\displaystyle R_{12}\ R_{13}\ R_{... | Wikipedia - YangβBaxter algebra | null | null | null |
Let A {\displaystyle A} be a unital commutative Banach algebra over C . {\displaystyle \mathbb {C} .} Since A {\displaystyle A} is then a commutative ring with unit, every non-invertible element of A {\displaystyle A} belongs to some maximal ideal of A . {\displaystyle A.} | Wikipedia - Banach ring | null | null | null |
Since a maximal ideal m {\displaystyle {\mathfrak {m}}} in A {\displaystyle A} is closed, A / m {\displaystyle A/{\mathfrak {m}}} is a Banach algebra that is a field, and it follows from the GelfandβMazur theorem that there is a bijection between the set of all maximal ideals of A {\displaystyle A} and the set Ξ ( A ) ... | Wikipedia - Banach ring | null | null | null |
A character Ο {\displaystyle \chi } is a linear functional on A {\displaystyle A} that is at the same time multiplicative, Ο ( a b ) = Ο ( a ) Ο ( b ) , {\displaystyle \chi (ab)=\chi (a)\chi (b),} and satisfies Ο ( 1 ) = 1. {\displaystyle \chi (\mathbf {1} )=1.} Every character is automatically continuous from A {\disp... | Wikipedia - Banach ring | null | null | null |
Moreover, the norm (that is, operator norm) of a character is one. Equipped with the topology of pointwise convergence on A {\displaystyle A} (that is, the topology induced by the weak-* topology of A β {\displaystyle A^{*}} ), the character space, Ξ ( A ) , {\displaystyle \Delta (A),} is a Hausdorff compact space. For... | Wikipedia - Banach ring | null | null | null |
{\displaystyle {\hat {x}}(\chi )=\chi (x).} The spectrum of x ^ , {\displaystyle {\hat {x}},} in the formula above, is the spectrum as element of the algebra C ( Ξ ( A ) ) {\displaystyle C(\Delta (A))} of complex continuous functions on the compact space Ξ ( A ) . | Wikipedia - Banach ring | null | null | null |
{\displaystyle \Delta (A).} Explicitly, As an algebra, a unital commutative Banach algebra is semisimple (that is, its Jacobson radical is zero) if and only if its Gelfand representation has trivial kernel. An important example of such an algebra is a commutative C*-algebra. In fact, when A {\displaystyle A} is a commu... | Wikipedia - Banach ring | null | null | null |
Let A {\displaystyle A} be an algebra and denote Con β‘ A {\displaystyle \operatorname {Con} A} the set of all congruences on A {\displaystyle A} . The set Con β‘ A {\displaystyle \operatorname {Con} A} is a complete lattice ordered by inclusion. If Ξ¦ β Con β‘ A {\displaystyle \Phi \in \operatorname {Con} A} is a congruen... | Wikipedia - Isomorphism theorems | null | null | null |
Let A {\displaystyle A} be an algebra and Ξ¦ , Ξ¨ {\displaystyle \Phi ,\Psi } two congruence relations on A {\displaystyle A} such that Ξ¨ β Ξ¦ {\displaystyle \Psi \subseteq \Phi } . Then Ξ¦ / Ξ¨ = { ( Ξ¨ , Ξ¨ ): ( a β² , a β³ ) β Ξ¦ } = Ξ¨ β Ξ¦ β Ξ¨ β 1 {\displaystyle \Phi /\Psi =\{(_{\Psi },_{\Psi }):(a',a'')\in \Phi \}=_{\Psi... | Wikipedia - First ring isomorphism theorem | null | null | null |
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