Extension of a bilinear form to the exterior algebra

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.

• Bourbaki, Algèbre 9, §1, no. 9.
– abx
Sep 17, 2019 at 4:10
• 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.
– YCor
Sep 17, 2019 at 10:39
• @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. Sep 18, 2019 at 4:16
• @YCor: You need alternatingness, not just multilinearity, and that's where the formula fails. Sep 18, 2019 at 5:28
• (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.)
– YCor
Sep 18, 2019 at 5:47

1 Answer

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.

• 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 $V=(v_1,v_2,\cdots,v_k),W=(w_1,w_2,\cdots,w_k).$ Sep 17, 2019 at 2:09
• @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). Sep 17, 2019 at 2:25