For $A$, $B$ and $C$ abelian groups and $A \times B$ the cartesian product group, a bilinear map
from $A$ and $B$ to $C$ is a function of the underlying sets (that is, a binary function from $A$ and $B$ to $C$) which is a linear map – that is a group homomorphism – in each argument separately.
In terms of elements this means that a bilinear map $f : A \times B \to C$ is a function of sets that satisfies for all elements $a_1, a_2 \in A$ and $b_1, b_2 \in B$ the two relations
and
Notice that this is not a group homomorphism out of the direct product group. The product group $A \times B$ is the group whose elements are pairs $(a,b)$ with $a \in A$ and $b \in B$, and whose group operation is
hence satisfies
and hence in particular
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:
For $A, B, C \in Ab$ a function of sets $f : A \times B \to C$ is a bilinear map from $A$ and $B$ to $C$ precisely if it factors through the tensor product of abelian groups $A \otimes B$ as
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.
More generally :
For $R$ a ring (or rig) and $A, B, C \in R$Mod being modules (say on the left, but on the right works similarly) over $R$, a bilinear map from $A$ and $B$ to $C$ is a function of the underlying sets
which is a bilinear map of the underlying abelian groups as in def. and in addition such that for all $r \in R$ we have
and
As before, this is equivalent to $f$ factoring through the tensor product of modules
Multilinear maps are again a generalisation.
See at tensor product of ∞-modules
A bilinear form is a bilinear map $f\colon A, B \to K$ whose target is the ground ring $K$; more generally, a multilinear form? is multilinear map whose target is $K$.
A bilinear map $f\colon A, A \to K$ whose two sources are the same is symmetric? if $f(a, b) = f(b, a)$ always; more generally, a multilinear map whose sources are all the same is symmetric? if $f(a_1, a_2, \ldots, a_n) = f(a_{\sigma(1)}, a_{\sigma(2)}, \ldots, a_{\sigma(n)})$ for each permutation $\sigma$ in the symmetric group $S_n$. (It's enough to check the $n-1$ generators of $S_n$ that transpose two adjacent arguments.) In particular, this defines symmetric bilinear and multilinear? forms.
A bilinear map $f\colon A, A \to K$ whose two sources are the same is antisymmetric? if $f(a, b) = -f(b, a)$ always; more generally, a multilinear map whose sources are all the same is antisymmetric? if $f(a_1, a_2, \ldots, a_n) = (-1)^\sigma f(a_{\sigma(1)}, a_{\sigma(2)}, \ldots, a_{\sigma(n)})$ for each permutation $\sigma$ in the symmetric group $S_n$, where $(-1)^\sigma$ is $1$ or $-1$ according as $\sigma$ is an even or odd permutation. (It's enough to check the $n-1$ generators of $S_n$ that transpose two adjacent arguments, which are each odd and so each introduce a factor of $-1$.) In particular, this defines antisymmetric bilinear? and multilinear? forms.
A bilinear map $f\colon A, A \to K$ whose two sources are the same is alternating? if $f(a, a) = 0$ always; more generally, a multilinear map whose sources are all the same is alternating if $f(a_1, a_2, \ldots, a_n) = 0$ whenever there exists a nontrivial permutation $\sigma$ in the symmetric group $S_n$ such that $(a_1, a_2, \ldots, a_n) = (a_{\sigma(1)}, a_{\sigma(2)}, \ldots, a_{\sigma(n)})$, in other words whenever there exist indexes $i \ne j$ such that $a_i = a_j$. (It's enough to say that $f(a_1, a_2, \ldots, a_n) = 0$ 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:
An alternating bilinear (or even multilinear) map must be antisymmetric.
If $f$ is an alternating bilinear map, then $f(a + b, a + b) = f(a, a) + f(a, b) + f(b, a) + f(b, b) = 0 + f(a, b) + f(b, a) + 0$, so $f(a, b) + f(b, a) = f(a + b, a + b) = 0$, so $f(a, b) = -f(b, a)$; that is, $f$ is antisymmetric. The general multilinear case is similar. (Note that linearity is essential to this proof.)
If the ground ring is a field whose characteristic is not $2$, or more generally if $1/2$ exists in the ground ring, or more generally if $2$ is cancellable in the target of the map in question, then an antisymmetric bilinear (or even multilinear) map must be alternating.
If $f$ is an antisymmetric bilinear map, then $f(a, a) = -f(a, a)$, so $2 f(a, a) = f(a, a) + f(a, a) = f(a, a) - f(a, a) = 0$, so $f(a, a) = 0$ (by dividing by $2$, multiplying by $1/2$, or cancelling $2$, 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 above:
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.
If $f$ is a trilinear map that alternates in each adjacent pair of arguments, then $f(a + b, a + b, a) = f(a, a, a) + f(a, b, a) + f(b, a, a) + f(b, b, a) = 0 + f(a, b, a) + 0 + 0$, so $f(a, b, a) = f(a + b, a + b, a) = 0$; that is, $f$ is alternating in the remaining pair of arguments. The general multilinear case is similar. (Again, linearity is essential to this proof.)
binary function, bilinear map, multilinear map
In the context of higher algebra/(∞,1)-category theory bilinear maps in an (∞,1)-category are discussed in section 4.3.4 of
Last revised on June 25, 2017 at 00:56:06. See the history of this page for a list of all contributions to it.