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\begin{document}

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\section*{bilinear map}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{group_theory}{}\paragraph*{{Group Theory}}\label{group_theory}

[[!include group theory - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak
\noindent\hyperlink{for_abelian_groups}{For abelian groups}\dotfill \pageref*{for_abelian_groups} \linebreak
\noindent\hyperlink{for_modules}{For modules}\dotfill \pageref*{for_modules} \linebreak
\noindent\hyperlink{ForInfinityModules}{For $\infty$-modules}\dotfill \pageref*{ForInfinityModules} \linebreak
\noindent\hyperlink{classification}{Classification}\dotfill \pageref*{classification} \linebreak
\noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{definition}{}\subsection*{{Definition}}\label{definition}

\hypertarget{for_abelian_groups}{}\subsubsection*{{For abelian groups}}\label{for_abelian_groups}

\begin{defn}
\label{BilinearOnAbelianGroups}\hypertarget{BilinearOnAbelianGroups}{}
For $A$, $B$ and $C$ [[abelian groups]] and $A \times B$ the [[cartesian product]] group, a \textbf{bilinear map}

\begin{displaymath}
f : A \times B \to C
\end{displaymath}
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.

\end{defn}
\begin{remark}
\label{}\hypertarget{}{}
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

\begin{displaymath}
f(a_1 + a_2, b_1) = f(a_1,b_1) + f(a_2, b_1)
\end{displaymath}
and

\begin{displaymath}
f(a_1, b_1 + b_2) = f(a_1, b_1) + f(a_1, b_2)
  \,.
\end{displaymath}
Notice that this is \emph{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

\begin{displaymath}
(a_1, b_1) + (a_2, b_2) = (a_1 + a_2 \;,\; b_1 + b_2)
  \,.
\end{displaymath}
A \emph{[[group homomorphism]]}

\begin{displaymath}
\phi : A \times B \to C
\end{displaymath}
hence satisfies

\begin{displaymath}
\phi( a_1+a_2, b_1 + b_2 ) = \phi(a_1,b_1) + \phi(a_2, b_2)
\end{displaymath}
and hence in particular

\begin{displaymath}
\phi( a_1+a_2, b_1  ) = \phi(a_1,b_1) + \phi(a_2, 0)
\end{displaymath}
\begin{displaymath}
\phi( a_1, b_1 + b_2 ) = \phi(a_1,b_1) + \phi(0, b_2)
\end{displaymath}
which is (in general) different from the behaviour of a bilinear map.

\end{remark}
The definition of [[tensor product of abelian groups]] is precisely such that the following is an equivalent definition of bilinear map:

\begin{defn}
\label{}\hypertarget{}{}
For $A, B, C \in Ab$ a function of sets $f : A \times B \to C$ is a \textbf{bilinear map} from $A$ and $B$ to $C$ precisely if it factors through the [[tensor product of abelian groups]] $A \otimes B$ as

\begin{displaymath}
f : A \times B \to A \otimes B \to C
  \,.
\end{displaymath}
\end{defn}
\begin{remark}
\label{}\hypertarget{}{}
The analogous defintion for more than two arguments yields \textbf{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]].

\end{remark}
\hypertarget{for_modules}{}\subsubsection*{{For modules}}\label{for_modules}

More generally :

\begin{defn}
\label{}\hypertarget{}{}
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 \textbf{bilinear map} from $A$ and $B$ to $C$ is a function of the underlying sets

\begin{displaymath}
f : A \times B \to C
\end{displaymath}
which is a bilinear map of the underlying [[abelian groups]] as in def. \ref{BilinearOnAbelianGroups} and in addition such that for all $r \in R$ we have

\begin{displaymath}
f(r a, b) = r f(a,b)
\end{displaymath}
and

\begin{displaymath}
f(a, r b) = r f(a,b)
  \,.
\end{displaymath}
\end{defn}
As before, if $R$ is [[commutative ring|commutative]] then this is equivalent to $f$ factoring through the [[tensor product of modules]]

\begin{displaymath}
f : A \times B \to A \otimes_R B \to C
  \,.
\end{displaymath}
\textbf{Multilinear maps} are again a generalisation.

\hypertarget{ForInfinityModules}{}\subsubsection*{{For $\infty$-modules}}\label{ForInfinityModules}

(\hyperlink{Lurie}{Lurie, section 4.3.4})

See at \emph{[[tensor product of ∞-modules]]}

\hypertarget{classification}{}\subsection*{{Classification}}\label{classification}

A \textbf{[[bilinear form]]} is a bilinear map $f\colon A, B \to K$ whose target is the [[ground ring]] $K$; more generally, a \textbf{[[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 \textbf{[[symmetric bilinear map|symmetric]]} if $f(a, b) = f(b, a)$ always; more generally, a multilinear map whose sources are all the same is \textbf{[[symmetric multilinear map|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 [[symmetric bilinear form|bilinear]] and [[symmetric multilinear form|multilinear]] forms.

A bilinear map $f\colon A, A \to K$ whose two sources are the same is \textbf{[[antisymmetric bilinear map|antisymmetric]]} if $f(a, b) = -f(b, a)$ always; more generally, a multilinear map whose sources are all the same is \textbf{[[antisymmetric multilinear map|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 [[antisymmetric bilinear form|bilinear]] and [[antisymmetric multilinear form|multilinear]] forms.

A bilinear map $f\colon A, A \to K$ whose two sources are the same is \textbf{[[alternating bilinear map|alternating]]} if $f(a, a) = 0$ always; more generally, a multilinear map whose sources are all the same is \textbf{[[alternating multilinear map|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 [[alternating bilinear form|bilinear]] and [[alternating multilinear form|multilinear]] forms.

In many cases, antisymmetric and alternating maps are equivalent:

\begin{prop}
\label{alternationImpliesAntisymmetry}\hypertarget{alternationImpliesAntisymmetry}{}
An alternating bilinear (or even multilinear) map must be antisymmetric.

\end{prop}
\begin{proof}
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.)

\end{proof}
\begin{prop}
\label{antisymmetryImpliesAlternation}\hypertarget{antisymmetryImpliesAlternation}{}
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.

\end{prop}
\begin{proof}
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.)

\end{proof}
The argument that the simplified description of alternation is correct is along the same lines as Proposition \ref{alternationImpliesAntisymmetry} above:

\begin{prop}
\label{transpositionsSufficeForAlternation}\hypertarget{transpositionsSufficeForAlternation}{}
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.

\end{prop}
\begin{proof}
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.)

\end{proof}
\hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts}

\begin{itemize}%
\item [[binary function]], \textbf{bilinear map}, \textbf{multilinear map}


\item [[binary morphism]], [[multimorphism]]


\item [[bifunctor]], [[Quillen bifunctor]]



\end{itemize}
\hypertarget{references}{}\subsection*{{References}}\label{references}

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

\begin{itemize}%
\item [[Jacob Lurie]], \emph{[[Higher Algebra]]}

\end{itemize}
[[!redirects bilinear map]] [[!redirects bilinear maps]] [[!redirects bilinear mapping]] [[!redirects bilinear mappings]] [[!redirects bilinear function]] [[!redirects bilinear functions]]

[[!redirects bilinear operator]] [[!redirects bilinear operators]]

[[!redirects multilinear map]] [[!redirects multilinear maps]] [[!redirects multilinear mapping]] [[!redirects multilinear mappings]] [[!redirects multilinear function]] [[!redirects multilinear functions]]

[[!redirects bilinear map in an (∞,1)-category]] [[!redirects bilinear maps in an (∞,1)-category]] [[!redirects bilinear map in an (infinity,1)-category]] [[!redirects bilinear maps in an (infinity,1)-category]]



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