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An exercise of my last year's linear algebra class asks as follows:

Determine the type of the quadratic form $q: \bigwedge^2 \mathbb{R}^4 \to \bigwedge^4 \mathbb{R}^4, x \mapsto x \wedge x$.

Question 1: We have only introduced quadratic forms of bilinear forms and I can't recall our professor talking about different "types" of quadratic forms. What types of quadratic forms are there?

Question 2: In the standard solution, one is adviced to proceed as follows: We have $\dim(\bigwedge^2 \mathbb{R}^4) = 6$ and thus the quadratic form will be represented by a 6x6-matrix no matter what basis we choose. We now choose the basis

$$\mathcal{B} = (e_1 \wedge e_2, e_1 \wedge e_3, e_1 \wedge e_4, e_2 \wedge e_3, e_2 \wedge e_4, e_3 \wedge e_4)$$

and by plugging $\mathcal{B}$ into the quadratic form, we arrive at the matrix

$$M_\mathcal{B}(q) = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}.$$

I don't understand how we arrive at this matrix. Can anyone explain?

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The matrix represents the associated symmetric bilinear form (with respect to your basis) $(,)$ defined by $$(u,v)=\frac12\left(q(u+v)-q(u)-q(v)\right)=u\wedge v.$$ –  Jyrki Lahtonen Aug 11 '12 at 20:15

1 Answer 1

up vote 4 down vote accepted

The only way to obtain a nontrivial 4-form in $\mathbb{R}^4$ is to have some scalar multiple of the top-form $e_1 \wedge e_2 \wedge e_3 \wedge e_4$. As you examine the basis you have provided you'll see that this top-form is only obtained via certain products.

Denoting $f_1 = e_1 \wedge e_2$ etc... $f_6 = e_3 \wedge e_4$ we calculate $f_1 \wedge f_6 = e_1 \wedge e_2 \wedge e_3 \wedge e_4$. On the other hand, $f_1 \wedge f_j=0$ for $j \neq 6$.

Continuing, examine $f_2 = e_1 \wedge e_3$ with $f_5=e_2 \wedge e_4$ gives the nontrivial $f_2 \wedge f_5 = e_1 \wedge e_3 \wedge e_2 \wedge e_4 = -e_1 \wedge e_2 \wedge e_3 \wedge e_4$ and in contrast, $f_2 \wedge f_j = 0$ for $j \neq 5$.

You can easily see that $f_3 \wedge f_4 = e_1 \wedge e_2 \wedge e_3 \wedge e_4$ and $f_3 \wedge f_j = 0$ for $j \neq 4$.

Since the wedge product of two forms commutes it follows that the results above capture all possible products. Let the matrix $M$ be defined implicitly via the equation $M_{ij}e_1 \wedge e_2 \wedge e_3 \wedge e_4 = f_i \wedge f_j$.

Abusing notation considerably, $Q(x) = x^TMx$ where I am replacing the two-form $x$ with it's column vector and identifying the top-form $e_1 \wedge e_2 \wedge e_3 \wedge e_4$ with $1$. As an example, $x = f_1+f_6 = [1,0,0,0,0,1]^T$ and we calculate, $$ Q(x) = [1,0,0,0,0,1]\left[ \begin{array}{cccccc} 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & -1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 1 \end{array} \right]\left[ \begin{array}{c} 1 \\ 0 \\0 \\0 \\ 0 \\ 1 \end{array} \right] = 2 $$ Which is precisely what we expect from direct computation of $Q$; $$(f_1+f_6) \wedge (f_1+f_6) = f_1 \wedge f_1 + f_1 \wedge f_6+f_6 \wedge f_1+ f_6 \wedge f_6 = 2e_1 \wedge e_2 \wedge e_3 \wedge e_4.$$

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