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.

  • 1
    $\begingroup$ Bourbaki, Algèbre 9, §1, no. 9. $\endgroup$
    – abx
    Sep 17, 2019 at 4:10
  • $\begingroup$ A bilinear form is "the same as" a homomorphism $f:V\to V^*$. Then it induces a homomorphism $\Lambda^k V\to \Lambda^k V^*$ by $f(v_1\wedge\dots \wedge v_k)=f(v_1)\wedge \dots \wedge f(v_k)$. Next one has a canonical homomorphism $\Lambda^k V^*\to (\Lambda^k V)^*$ given by $(L_1\wedge \dots\wedge L_k)\mapsto (v_1\wedge\dots\wedge v_k\mapsto L_1(v_1)\wedge\dots L_k(v_k))$. Here I seem to assume to work in a field, but makes sense in a module context if $V^*$ is defined as the module of homomorphisms into a fixed module $E$, if we consider forms as being $E$-valued. $\endgroup$
    – YCor
    Sep 17, 2019 at 10:39
  • $\begingroup$ @YCor: Be careful with the definition of your homomorphism $\Lambda^k V^\ast \to \left(\Lambda^k V\right)^\ast$; the formula you give doesn't quite make sense (its right hand side depends on the way the left hand side is represented). Good idea, though. $\endgroup$ Sep 18, 2019 at 4:16
  • $\begingroup$ @YCor: You need alternatingness, not just multilinearity, and that's where the formula fails. $\endgroup$ Sep 18, 2019 at 5:28
  • $\begingroup$ (It made sense, but let to a zero map since valued in $\Lambda^k$ of a 1-dimensional space for $k\ge 2$, so indeed wrong.) $\endgroup$
    – YCor
    Sep 18, 2019 at 5:47

1 Answer 1


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). 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.

  • $\begingroup$ Thanks for your comprehensive answer. Lets consider the following simple case: let $V$ be a $k$-vector space of finite dimension. After choosing a basis, we can identify $V$ with $k^n$. $T$ is given by a $n\times n$ matrix: $T(v,w)=v^T Mw$. So the $\alpha_k$ above is simply given by the following formula:$$\alpha_k(v_1\wedge v_2\wedge\cdots\wedge v_k,w_1\wedge w_2\wedge\cdots\wedge w_k)=\textrm{det}(V^TMW),$$ where $V=(v_1,v_2,\cdots,v_k),W=(w_1,w_2,\cdots,w_k).$ $\endgroup$
    – tanjia
    Sep 17, 2019 at 2:09
  • $\begingroup$ @tanjia: Yes, that's a nice way to put it! This also makes me think of an easier way to prove non-degeneracy (I'm editing my post). $\endgroup$ Sep 17, 2019 at 2:25

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