# How to find matrix of antisymmetrization $\pi_A(g)$ where $g$ is the bilinear form $e^1\otimes e^1-e^1\otimes e^2+3e^2\otimes e^1+2e^2\otimes e^2$

Basis $$M=\{(3,1),(2,1)\}$$. I solved that the dual basis $$\begin{equation} M^*=\{e^1,e^2\}=\{(1,-2),(-1,3)\} \end{equation}$$

Then I solved that matrix

$$\begin{equation} g=\left(\begin{array}{cc}1&-5\\-1&10\end{array}\right) \end{equation}$$

Now, my notes says that antisymmetrization matrix for purely covariant tensor (which is a bilinear form) is given as

$$\frac{1}{q!}\sum_{\pi \in S_q}\text{sgn}\pi.(e_{\pi(1)},.....,e_{\pi(n)})$$

Here $$q$$ is 2 (because bilinear form is a (0,2) tensor and since we're in 2 dimensional space, there should be two permutations).

If I understant correctly, my matrix should be equal to

$$\frac{1}{2}\left((e_1,e_2)-(e_2,e_1)\right),$$

I just don't know what these $$e_1$$ and $$e_2$$ mean (I suppose they are not meant to be the same as the basis vectors of $$M$$).

Are $$e_1$$ and $$e_2$$ meant to be columns of the matrix $$g$$?

In that case the matrix of antisymmetrization would be

$$\pi_A(g)=\frac{1}{2}\left(\begin{array}{cc}6&-6\\-11&11\end{array}\right)$$

Actually, the $$e_{\pi(i)}$$ are inteded to be the same you computed in the dual basis, but with another meaning. In fact, $$(e_2,e_1)$$ and $$(e_1,e_2)$$ are to be intended as a permutation operators on the tensor.
In particular, $$(e_1,e_2)$$ is the identity operator, whereas $$(e_2,e_1)$$ swaps the indices as $$(e_2,e_1) (e_1\otimes e_1) =e_2 \otimes e_2,\\ (e_2,e_1) (e_1\otimes e_2) =e_2 \otimes e_1,\\\dots$$ Notice that these operators are linear, and you can compute their effect on the matrix. In fact, $$(e_1,e_2)$$ does not change the matrix $$g$$, whereas $$(e_2,e_1)$$ coincides with the transposition of the matrix.
In your case, $$\pi_A(g)=\frac{1}{2}(g-g^T)$$ that is what is commonly known as the skew(or anti)-symmetric part of $$g$$.