In Serre's *Local Fields*, at the beginning of the chapter III section 2, he has wrote "it is known that $T$ extends to a non-degenerate bilinear form on the exterior algebra of $V$", where $T$ is a non-degenerated bilinear form over a vector space $V$.


I get confused about this well-know extension. Is there any explicit formula for this extension? Thank you.



Extension of a bilinear form to the exterior algebra
Let $k$ be a nonnegative integer. Let $K$ be a commutative ring, and let $V$ and $W$ be two $K$-modules. Let $\alpha : V \times W \to K$ be a $K$-bilinear form. Then, there is a $K$-bilinear form
\begin{align}
\alpha\_k : \wedge^k V \times \wedge^k W &\to K; \\
\left(v\_1 \wedge v\_2 \wedge \cdots \wedge v\_k , w\_1 \wedge w\_2 \wedge \cdots \wedge w\_k\right) &\mapsto \det\left(\left( \alpha\left(v\_i, w\_j\right) \right)\_{1\leq i\leq k, \ 1\leq j\leq k}\right)
\end{align}
(where $\left( \alpha\left(v\_i, w\_j\right) \right)\_{1\leq i\leq k, \ 1\leq j\leq k}$ is the $k\times k$-matrix with entries $\alpha\left(v\_i, w\_j\right)$).
Note that the image of $\left(v\_1 \wedge v\_2 \wedge \cdots \wedge v\_k , w\_1 \wedge w\_2 \wedge \cdots \wedge w\_k\right)$ under this form is often called a Gram determinant (though the name traditionally stands for some particular cases), and can be rewritten as $\sum\_{\sigma \in S\_k} \left(-1\right)^{\sigma} \prod\_{i=1}^k \alpha\left(v\_i, w\_{\sigma\left(i\right)}\right)$ (where $S\_k$ is the symmetric group on the set $\left\{1,2,\ldots,k\right\}$, and where $\left(-1\right)^{\sigma}$ denotes the sign of a permutation $\sigma$).


Why is this form $\alpha\_k$ well-defined? Here is the straightforward way to see this (there might be slicker arguments): For each $v\_1, v\_2, \ldots, v\_k \in V$, we define a $K$-linear map
\begin{align}
A\_{v\_1, v\_2, \ldots, v\_k} : \wedge^k W &\to K; \\
w\_1 \wedge w\_2 \wedge \cdots \wedge w\_k &\mapsto \det\left(\left( \alpha\left(v\_i, w\_j\right) \right)\_{1\leq i\leq k, \ 1\leq j\leq k}\right) .
\end{align}
This is well-defined (by the universal property of $\wedge^k W$), since the determinant $\det\left(\left( \alpha\left(v\_i, w\_j\right) \right)\_{1\leq i\leq k, \ 1\leq j\leq k}\right)$ depends $K$-multilinearly on the $w\_1, w\_2, \ldots, w\_k$ and vanishes when two of $w\_1, w\_2, \ldots, w\_k$ are equal (indeed, this follows from standard properties of determinants). Now, consider the map
\begin{align}
A : V^{\times k} &\to \operatorname{Hom}\_K\left(\wedge^k W, K\right); \\
\left(v\_1, v\_2, \ldots, v\_k\right) &\mapsto A\_{v\_1, v\_2, \ldots, v\_k} .
\end{align}
This map $A$ is $K$-multilinear and kills all $k$-tuples $\left(v\_1, v\_2, \ldots, v\_k\right) \in V^{\times k}$ that have two equal entries (indeed, this all boils down to checking identities between $K$-linear maps, such as
\begin{align}
A\_{v\_1, v\_2, \ldots, v\_{i-1}, v\_i + v'\_i, v\_{i+1}, \ldots, v\_k}
&= A\_{v\_1, v\_2, \ldots, v\_{i-1}, v\_i, v\_{i+1}, \ldots, v\_k}
+ A\_{v\_1, v\_2, \ldots, v\_{i-1}, v'\_i, v\_{i+1}, \ldots, v\_k} ; \\
A\_{v\_1, v\_2, \ldots, v\_{i-1}, \lambda v\_i, v\_{i+1}, \ldots, v\_k}
&= \lambda A\_{v\_1, v\_2, \ldots, v\_{i-1}, v\_i, v\_{i+1}, \ldots, v\_k} ; \\
A\_{v\_1, v\_2, \ldots, v\_{i-1}, v, v, v\_{i+2}, \ldots, v\_k}
&= 0 ;
\end{align}
but any identity between $K$-linear maps can be proven by evaluating both sides at the generators $w\_1 \wedge w\_2 \wedge \cdots \wedge w\_k$ of $\wedge^k W$; but when evaluated this way, all these identities again follow from basic properties of the determinant). Thus, this map $A$ gives rise to a $K$-linear map
\begin{align}
A' : \wedge^k V &\to \operatorname{Hom}\_K\left(\wedge^k W, K\right); \\
v\_1 \wedge v\_2 \wedge \cdots \wedge v\_k &\mapsto A\_{v\_1, v\_2, \ldots, v\_k}
\end{align}
 (by the universal property of $\wedge^k V$).
This map $A'$, in turn, induces a $K$-bilinear form
\begin{align}
\alpha\_k : \wedge^k V \times \wedge^k W &\to K; \\
\left(p, q\right) &\mapsto \left(A'\left(p\right)\right)\left(q\right)
\end{align}
(by [uncurrying](https://en.wikipedia.org/wiki/Currying)). An immediate verification reveals that this $K$-bilinear form $\alpha\_k$ is precisely the form $\alpha\_k$ presented above.


Of course, by considering $k$ as variable, we can glue these $K$-bilinear forms $\alpha\_k : \wedge^k V \times \wedge^k W \to K$ together into a $K$-bilinear form $\alpha\_\wedge : \wedge V \times \wedge W \to K$. This latter form, I believe, is the form Serre wants.


Why is this latter form $\alpha\_\wedge$ non-degenerate when $\alpha$ is non-degenerate? Here, we say that a $K$-bilinear form $\beta : P \times Q \to K$ is non-degenerate if there exist bases $\left(p\_i\right)\_{i \in I}$ and $\left(q\_i\right)\_{i \in I}$ of $P$ and $Q$, respectively, such that $\beta\left(p\_i, q\_j\right) = \delta\_{i, j}$ for all $i \in I$ and $j \in I$. Two such bases are called *dual bases* for the form $\beta$.


Now, assume that $\alpha$ is non-degenerate. Thus, there exist dual bases $\left(v\_i\right)\_{i \in I}$ and $\left(w\_i\right)\_{i \in I}$ for the form $\alpha$. Consider such bases. WLOG assume that the set $I$ is totally ordered (since we can always equip $I$ with a total order). The basis $\left(v\_i\right)\_{i \in I}$ of $V$ induces a basis $\left(v\_{i\_1} \wedge v\_{i\_2} \wedge \cdots \wedge v\_{i\_k}\right)\_{\left(i\_1, i\_2, \ldots, i\_k\right) \in I\_k}$ of $\wedge^k V$, where $I\_k$ is the set of all strictly increasing $k$-tuples $\left(i\_1 < i\_2 < \cdots < i\_k\right) \in I^k$. Similarly, the basis $\left(w\_i\right)\_{i \in I}$ of $W$ induces a basis $\left(w\_{i\_1} \wedge w\_{i\_2} \wedge \cdots \wedge w\_{i\_k}\right)\_{\left(i\_1, i\_2, \ldots, i\_k\right) \in I\_k}$ of $\wedge^k W$. It is now easy to see that these two bases of $\wedge^k V$ and $\wedge^k W$ are dual bases for the $K$-bilinear form $\alpha\_k$. Thus, the $K$-bilinear form $\alpha\_k$ has dual bases, i.e., is non-degenerate. Qed.


