# Identification between wedge product and its dual

Let $\mathbb{F}$ be a field, and let $(e_i)$ be the usual elementary basis of $\mathbb{F}^n$. Let $\varphi_{ij}: \mathbb{F}^n \wedge \mathbb{F}^n \to \mathbb{F}$ be such that $v \wedge w \mapsto v_iw_j - w_iv_j$. It is not hard to show that the $\varphi_{ij}$ are the dual basis of $(e_i \wedge e_j)$.

Is there a canonical reason to identify $\varphi_{ij}$ with $e_i^* \wedge e_j^*$, or does this amount to just a arbitrary choice of isomorphism $(\mathbb{F}^n)^* \wedge (\mathbb{F}^n)^* \to (\mathbb{F}^n \wedge \mathbb{F}^n)^*$? That is, if we are in the exterior algebra of $(\mathbb{F}^n)^*$ and we take the wedge product of $e_i^*$ and $e_j^*$, why should we end up with $\varphi_{ij}$?

• For any $\mathbb{F}$-vector space $V$, there is a bilinear form $\left(\wedge^2\left(V^\ast\right)\right) \times \left(\wedge^2 V\right) \to \mathbb{F},\ \left(\left(f \wedge g\right),\left(a \wedge b\right)\right) \mapsto f\left(a\right)g\left(b\right)-g\left(a\right)f\left(b\right)$. (This actually generalizes to higher exterior powers; the form then returns a determinant. I am sure it has a name.) This bilinear form is nondegenerate, and thus gives rise to an isomorphism $\wedge^2 \left(V^\ast\right) \to \left(\wedge^2 V\right)^\ast$, which becomes your isomorphism when $V = \mathbb{F}^n$. – darij grinberg Jul 28 '15 at 21:35
• Note that your title is imprecise: it's not its dual. – darij grinberg Jul 28 '15 at 21:36

Let me reformulate the question. You are essentially asking if the isomorphism $\Lambda^2(V^*) \rightarrow (\Lambda^2 V)^*$ where $\varphi^* \wedge \psi^* \mapsto [v \wedge w \mapsto \varphi(v)\psi(w) - \psi(v)\varphi(w)]$ is "canonically" induced by the natural isomorphism $$V^* \otimes V^* \rightarrow (V \otimes V)^*$$ where $\varphi \otimes \psi \mapsto [v \otimes w \mapsto \varphi(v)\psi(w)]$ by somehow collapsing both sides of this isomorphism into stuff involving exterior powers: the left side into $\Lambda^2(V^*)$ and the right side into $(\Lambda^2 V)^*$.

An exterior power is canonically (and I'd say even is by definition) a quotient of a tensor power: there is a linear surjection $V \otimes V \rightarrow \Lambda^2(V)$ by $v \otimes w \mapsto v \wedge w$ and also a linear surjection $V^* \otimes V^* \rightarrow \Lambda^2(V^*)$ by $\varphi \otimes \psi \mapsto \varphi \wedge \psi$. I think you are seeking a commutative diagram

$$\begin{array}[c]{ccc} V^* \otimes V^* &{\rightarrow}&(V \otimes V)^*\\ \downarrow &&\downarrow \\ \Lambda^2(V^*)&{\rightarrow}&(\Lambda^2 V)^* \end{array}$$ where the horizontal maps are the isomorphisms described above and the map on the left is the canonical quotient map defining $\Lambda^2(V^*)$ as a quotient of $V^* \otimes V^*$. What map could fit on the right?

Since $\Lambda^2(V)$ is a quotient of $V \otimes V$, the map of dual spaces on the right can't be the dual of the quotient map $V \otimes V \rightarrow \Lambda^2(V)$ since it would go the wrong way. To write down a map $(V \otimes V)^* \rightarrow (\Lambda^2 V)^*$ we'd like a linear map $\Lambda^2(V) \rightarrow V \otimes V$ that we can dualize. There is a standard linear map $\alpha_{2,V} \colon \Lambda^2(V) \rightarrow V \otimes V$, given by $v \wedge w \mapsto v \otimes w - w \otimes v$ on elementary wedge products, it is injective (for all scalar fields), and using its dual map $\alpha_{2,V}^* : (V \otimes V)^* \rightarrow (\Lambda^2 V)^*$ on the right in the diagram above makes that diagram commute: check what happens when you take an elementary tensor in the upper left around both ways to the lower right. You'll get the same result both ways.

More generally, for any commutative ring $R$, $R$-module $M$, and integer $k \geq 1$ there is a canonical quotient map $M^{\otimes k} \rightarrow \Lambda^k(M)$ where $m_1 \otimes \cdots \otimes m_k \mapsto m_1 \wedge \cdots \wedge m_k$ on elementary tensors and also a linear map the other way $\alpha_{k,M} \colon \Lambda^k(M) \rightarrow M^{\otimes k}$ where $$\alpha_{k,M}(m_1 \wedge \cdots \wedge m_k) = \sum_{\sigma \in S_k} ({\rm sign}\,\sigma) m_{\sigma(1)} \wedge \cdots \wedge m_{\sigma(k)}$$ on elementary wedge products. (Warning: the composite of these maps $\Lambda^k(M) \rightarrow M^{\otimes k} \rightarrow \Lambda^k(M)$ is not the identity, but is multiplication by $k!$.) Using these two linear maps we get a commutative digram of $R$-modules and $R$-linear maps $$\begin{array}[c]{ccc} (M^*)^{\otimes k} &{\rightarrow}&(M^{\otimes k})^*\\ \downarrow &&\downarrow \\ \Lambda^k(M^*)&{\rightarrow}&(\Lambda^k M)^* \end{array}$$ generalizing the commutative diagram above ($M = V$, $k = 2$), where the map on the top is canonical, having the effect $f_1 \otimes \cdots \otimes f_k \mapsto [m_1 \otimes \cdots \otimes m_k \mapsto f_1(m_1) \cdots f_k(m_k)]$ on elementary tensors, the linear map on the left is the canonical quotient map, the linear map on the right is the dual of $\alpha_{k,M}$, and the linear map on the bottom has the effect $f_1 \wedge \cdots \wedge f_k \mapsto [m_1 \wedge\cdots \wedge m_k \mapsto \det(f_i(m_j))]$ on elementary wedge products. In this generality the horizontal maps need not be isomorphisms, unlike in the case of finite-dimensional vector spaces, but the diagram still commutes.

The main observation to make here is that the map along the bottom, which is what your question is asking above, should not be considered compatible just with the natural maps on the top and left, but also with the map along the right side. The maps along the bottom and the right both involve determinant-looking expressions (sums over the symmetric group with terms permuted according to the sign of a permutation and scaled by the sign of that permutation), and they work well together by giving us the above commutative diagram. The map along the top doesn't on its own induce the map on the bottom because there is no obvious map from $(M^{\otimes k})^*$ to $(\Lambda^k M)^*$ using only the quotient map $M^{\otimes k} \rightarrow \Lambda^k M$, since the dual of this quotient map is going the wrong way. We put determinant-type stuff into the situation not only along the bottom of the commutative digram, but also along the right, and you need to have both of them in mind.

• I note that one convenience of this choice of right/bottom maps is that if $v_1, \dotsc, v_k$ are elements of a basis for $M$, and $v_1^*, \dotsc, v_k^*$ are their duals with respect to this basis, then $v_{1}^* \wedge \dotsb \wedge v_{k}^* \mapsto (v_1 \wedge \dotsb \wedge v_k)^*$. – Eric Auld Aug 31 '15 at 18:37
• It is also characterized by this property in the case of finite-dimensional vector spaces. I guess this is another thing that makes the identification so natural (and why most authors gloss over it). – Eric Auld Sep 1 '15 at 5:52

For any two vector spaces $V$ and $W$ over the same field, there is a canonical isomorphism $V^*\otimes W^* \cong (V\otimes W)^*$, given by $(a\otimes b)(v\otimes w) = a(v) b(w)$, extended by linearity.

This is just a special case, where $W = V$ and we anti-symmetrise.