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bilinear map

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Definition

For abelian groups

Definition

For AA, BB and CC abelian groups and A×BA \times B the cartesian product group, a bilinear map

f:A×BC f : A \times B \to C

from AA and BB to CC is a function of the underlying sets (that is, a binary function from AA and BB to CC) which is a linear map – that is a group homomorphism – in each argument separately.

Remark

In terms of elements this means that a bilinear map f:A×BCf : A \times B \to C is a function of sets that satisfies for all elements a 1,a 2Aa_1, a_2 \in A and b 1,b 2Bb_1, b_2 \in B the two relations

f(a 1+a 2,b 1)=f(a 1,b 1)+f(a 2,b 1) f(a_1 + a_2, b_1) = f(a_1,b_1) + f(a_2, b_1)

and

f(a 1,b 1+b 2)=f(a 1,b 1)+f(a 1,b 2). f(a_1, b_1 + b_2) = f(a_1, b_1) + f(a_1, b_2) \,.

Notice that this is not a group homomorphism out of the direct product group. The product group A×BA \times B is the group whose elements are pairs (a,b)(a,b) with aAa \in A and bBb \in B, and whose group operation is

(a 1,b 1)+(a 2,b 2)=(a 1+a 2,b 1+b 2). (a_1, b_1) + (a_2, b_2) = (a_1 + a_2 \;,\; b_1 + b_2) \,.

A group homomorphism

ϕ:A×BC \phi : A \times B \to C

hence satisfies

ϕ(a 1+a 2,b 1+b 2)=ϕ(a 1,b 1)+ϕ(a 2,b 2) \phi( a_1+a_2, b_1 + b_2 ) = \phi(a_1,b_1) + \phi(a_2, b_2)

and hence in particular

ϕ(a 1+a 2,b 1)=ϕ(a 1,b 1)+ϕ(a 2,0) \phi( a_1+a_2, b_1 ) = \phi(a_1,b_1) + \phi(a_2, 0)
ϕ(a 1,b 1+b 2)=ϕ(a 1,b 1)+ϕ(0,b 2) \phi( a_1, b_1 + b_2 ) = \phi(a_1,b_1) + \phi(0, b_2)

which is (in general) different from the behaviour of a bilinear map.

The definition of tensor product of abelian groups is precisely such that the following is an equivalent definition of bilinear map:

Definition

For A,B,CAbA, B, C \in Ab a function of sets f:A×BCf : A \times B \to C is a bilinear map from AA and BB to CC precisely if it factors through the tensor product of abelian groups ABA \otimes B as

f:A×BABC. f : A \times B \to A \otimes B \to C \,.
Remark

The analogous defintion for more than two arguments yields multilinear maps. There is a multicategory of abelian groups and multilinear maps between them; the bilinear maps are the binary morphisms, and the multilinear maps are the multimorphisms.

For modules

More generally :

Definition

For RR a ring (or rig) and A,B,CRA, B, C \in RMod being modules (say on the left, but on the right works similarly) over RR, a bilinear map from AA and BB to CC is a function of the underlying sets

f:A×BC f : A \times B \to C

which is a bilinear map of the underlying abelian groups as in def. 1 and in addition such that for all rRr \in R we have

f(ra,b)=rf(a,b) f(r a, b) = r f(a,b)

and

f(a,rb)=rf(a,b). f(a, r b) = r f(a,b) \,.

As before, this is equivalent to ff factoring through the tensor product of modules

f:A×BA RBC. f : A \times B \to A \otimes_R B \to C \,.

Multilinear maps are again a generalisation.

For \infty-modules

(Lurie, section 4.3.4)

See at tensor product of ∞-modules

Classification

A bilinear form is a bilinear map f:A,BKf\colon A, B \to K whose target is the ground ring KK; more generally, a multilinear form? is multilinear map whose target is KK.

A bilinear map f:A,AKf\colon A, A \to K whose two sources are the same is symmetric? if f(a,b)=f(b,a)f(a, b) = f(b, a) always; more generally, a multilinear map whose sources are all the same is symmetric? if f(a 1,a 2,,a n)=f(a σ(1),a σ(2),,a σ(n))f(a_1, a_2, \ldots, a_n) = f(a_{\sigma(1)}, a_{\sigma(2)}, \ldots, a_{\sigma(n)}) for each permutation σ\sigma in the symmetric group S nS_n. (It's enough to check the n1n-1 generators of S nS_n that transpose two adjacent arguments.) In particular, this defines symmetric bilinear and multilinear? forms.

A bilinear map f:A,AKf\colon A, A \to K whose two sources are the same is antisymmetric? if f(a,b)=f(b,a)f(a, b) = -f(b, a) always; more generally, a multilinear map whose sources are all the same is antisymmetric? if f(a 1,a 2,,a n)=(1) σf(a σ(1),a σ(2),,a σ(n))f(a_1, a_2, \ldots, a_n) = (-1)^\sigma f(a_{\sigma(1)}, a_{\sigma(2)}, \ldots, a_{\sigma(n)}) for each permutation σ\sigma in the symmetric group S nS_n, where (1) σ(-1)^\sigma is 11 or 1-1 according as σ\sigma is an even or odd permutation. (It's enough to check the n1n-1 generators of S nS_n that transpose two adjacent arguments, which are each odd and so each introduce a factor of 1-1.) In particular, this defines antisymmetric bilinear? and multilinear? forms.

A bilinear map f:A,AKf\colon A, A \to K whose two sources are the same is alternating? if f(a,a)=0f(a, a) = 0 always; more generally, a multilinear map whose sources are all the same is alternating if f(a 1,a 2,,a n)=0f(a_1, a_2, \ldots, a_n) = 0 whenever there exists a nontrivial permutation σ\sigma in the symmetric group S nS_n such that (a 1,a 2,,a n)=(a σ(1),a σ(2),,a σ(n))(a_1, a_2, \ldots, a_n) = (a_{\sigma(1)}, a_{\sigma(2)}, \ldots, a_{\sigma(n)}), in other words whenever there exist indexes iji \ne j such that a i=a ja_i = a_j. (It's enough to say that f(a 1,a 2,,a n)=0f(a_1, a_2, \ldots, a_n) = 0 whenever two adjacent arguments are equal, although this is not as trivial as the analogous statements in the previous two paragraphs.) In particular, this defines alternating bilinear? and multilinear? forms.

In many cases, antisymmetric and alternating maps are equivalent:

Proposition

An alternating bilinear (or even multilinear) map must be antisymmetric.

Proof

If ff is an alternating bilinear map, then f(a+b,a+b)=f(a,a)+f(a,b)+f(b,a)+f(b,b)=0+f(a,b)+f(b,a)+0f(a + b, a + b) = f(a, a) + f(a, b) + f(b, a) + f(b, b) = 0 + f(a, b) + f(b, a) + 0, so f(a,b)+f(b,a)=f(a+b,a+b)=0f(a, b) + f(b, a) = f(a + b, a + b) = 0, so f(a,b)=f(b,a)f(a, b) = -f(b, a); that is, ff is antisymmetric. The general multilinear case is similar. (Note that linearity is essential to this proof.)

Proposition

If the ground ring is a field whose characteristic is not 22, or more generally if 1/21/2 exists in the ground ring, or more generally if 22 is cancellable in the target of the map in question, then an antisymmetric bilinear (or even multilinear) map must be alternating.

Proof

If ff is an antisymmetric bilinear map, then f(a,a)=f(a,a)f(a, a) = -f(a, a), so 2f(a,a)=f(a,a)+f(a,a)=f(a,a)f(a,a)=02 f(a, a) = f(a, a) + f(a, a) = f(a, a) - f(a, a) = 0, so f(a,a)=0f(a, a) = 0 (by dividing by 22, multiplying by 1/21/2, or cancelling 22, as applicable). The general multilinear case is similar. (Note that linearity is irrelevant to this proof.)

The argument that the simplified description of alternation is correct is along the same lines as Proposition 1 above:

Proposition

If a trilinear map is alternating in the first two arguments and in the last two arguments, or more generally if a multilinear map is alternating in every pair of adjacent arguments (or indeed in any set of transpositions that generate the entire symmetric group), then the map is alternating overall.

Proof

If ff is a trilinear map that alternates in each adjacent pair of arguments, then f(a+b,a+b,a)=f(a,a,a)+f(a,b,a)+f(b,a,a)+f(b,b,a)=0+f(a,b,a)+0+0f(a + b, a + b, a) = f(a, a, a) + f(a, b, a) + f(b, a, a) + f(b, b, a) = 0 + f(a, b, a) + 0 + 0, so f(a,b,a)=f(a+b,a+b,a)=0f(a, b, a) = f(a + b, a + b, a) = 0; that is, ff is alternating in the remaining pair of arguments. The general multilinear case is similar. (Again, linearity is essential to this proof.)

References

In the context of higher algebra/(∞,1)-category theory bilinear maps in an (∞,1)-category are discussed in section 4.3.4 of

Revised on June 25, 2017 00:56:06 by Toby Bartels (2605:6000:2b44:1100:fcfb:8220:11b9:95bf)