# How to identifiy $V \wedge V$ with the space of all alternating bilinear forms

Let $\{ e_i \}$ be a basis for $V$, then the space of tensors $V \otimes V$ could be identified with the space of all formal sums $\sum_{ij} \alpha_{ij} (e_i, e_j)$ (I know a base independent approach would be preferable, but I want to keep it short).

In the finite-dimensional setting each tensor $T = \sum_{ij} \alpha_{ij} (e_i \times e_j)$ could be identified with a bilinear map $T(u,v) = \sum_{ij} \alpha_{ij} \cdot e_i^*(u) \cdot e_j^*(v)$, where $e_i^*$ denotes the dual basis element defined by $e_i^*(e_j) := \delta_{ij}$, vice versa each bilinear map $T : U \times V \to K$ could be written as a tensor, because the maps $\{ \phi_{ij} \}$ with $\phi_{ij}(u,v) := e_i^*(u)\cdot e_j^*(u,v)$ form a basis of the space of all bilinear maps.

Now for a vector space $V$ set $A(V) := \mbox{span}\{ v \otimes v : v \in V \}$ and define $V \wedge V := V \otimes V / A(V)$. Now I read that $V \wedge V$ could be identified with the space of all alternating forms $V \times V \to K$ (and in many differential geometry books it is defined that way). But how is this identification achieved?

Now I computed an example for $V \wedge V$, let $V = \mathbb R^2$ and let $e_1, e_2$ be the standard bases, then a basis of $V \otimes V$ is $(e_1, e_1), (e_1, e_2), (e_2, e_1), (e_2, e_2)$. And for $v = \alpha e_1 + \beta e_2$ we have $$v \otimes v = \alpha^2 (e_1, e_1) + \alpha \beta (e_1, e_2) + \beta \alpha (e_2,e_1) + \beta^2 (e_2, e_2)$$ looking at the coordinates, I found that $A(V)$ equals the space where the second and third coordinates equals, i.e. we have $$A(V) = \mbox{span}\{ (e_1, e_1), (e_2, e_2), (e_1, e_2) + (e_2, e_1) \}$$ and these vector form a basis. With the above interpretation I found that $A(V)$ equals the space of all symmetric bilinear forms.

Now consider $V \wedge V = V \otimes V / A(V)$, then this space is $1$-dimensional, and because $(e_1, e_2) \notin A(V)$ we have $$V \wedge V = \mbox{span}\{ (e_1, e_2) + A(V) \}.$$

So now in what sense could $V \wedge V$ be identified with the space of all alternating bilinear maps? For example when I try an approach similar as the above, I would identify $\alpha(e_1, e_2) + v$ and $v = \beta(e_1, e_1) + \gamma(e_2, e_2) + \delta( (e_1,e_2) + (e_2, e_1) ) \in A(V)$ with $$\varphi(u,v) = \alpha e_1^*(u)e_2^*(v) + \beta e_1^*(u)e_1^*(v) + \gamma e_2^*(u)e_2^*(v) + \delta e_1^*(u)e_2^*(v) + \delta e_2^*(u)e_1^*(v).$$ But then not every such element gives an alternating form, for example $(e_1, e_2) + v$ itself, the map $\varphi(u,v) = e_1^*(u) \cdot e_2^*(v)$ is not alternating because $\varphi(e_1, e_2) = 1$, but $\varphi(e_2, e_1) = 0$.

So what went wrong here, and how could this identification achieved?

• for dimension two there is only one base alternating bilinear map which is the determinant of 2x2 matrices, so any other bilinear map is a scalar multiple of this. Oct 24, 2014 at 2:06
• I know that this space must have dimension $1$ in this case (and dimension $\binom{n}{2}$ in general), I asked about the specific identification and how is it related to my example/computation. Oct 24, 2014 at 9:01
• Ok.. in the other hand you can get alternating bilinear maps if you choose the components $\alpha_{ij}$ satisfying the antisymmetry condition i.e. $\alpha_{ij}=-\alpha_{ji}$ and choosing as basis for the subspace generated by $e^*_i\wedge e^*_j:=e^*_i\otimes e^*_j-e^*_j\otimes e^*_i$. Observe that $e^*_i\wedge e^*_j(v,w)=v_iw_j-v_jw_i$ Oct 25, 2014 at 19:43

The main trick here is

$$(x+y) \otimes (x+y) - x \otimes x - y \otimes y = x \otimes y + y \otimes x$$

I bet you're making things much harder by trying to do things in terms of a basis, rather than using tensor algebra.

Also, you're probably contributing to your confusion by identifying elements of $V \otimes V$ with as bilinear forms on $V$: it would be better to identify them with bilinear forms on $V^*$, or consider $V^* \otimes V^*$ instead.

• How is this "main trick" applied, sorry I do not understand... Oct 22, 2014 at 22:18
• @Stefan: On the one hand, I had misread your posting, and was trying to hint at something related to my misreading. But on the other hand, recognizing that $\wedge$ is antisymmetric is useful anyways. You can actually find an injective homomorphism $V \wedge V \to V \otimes V$. Either way, I think my other comments still apply.
– user14972
Oct 22, 2014 at 23:22
• Surely is there an injection $V \wedge V \to V \otimes V$, because it is a quotient, every representation system would give such an injection... but in what way is this related to my question, what have my construction's to to with the correspondence between $V \wedge V$ and the space of alternating linear forms? Oct 22, 2014 at 23:38
• But, it is of interest consider elements $Q$ of $V\otimes V$ as a bilinear form $V\times V\to \mathbb{F}$ defined via $(v,w)\mapsto v^{\top}[Q]w$, where $[Q]$ is the matrix $[Q(b_i,b_j)]$ and the $b_i$ basis for $V$. Nov 20, 2014 at 17:25