Coordinates of skew symmetric tensors

Question

Let $T$ be a tensor of type $(p,0)$, i.e. we consider multilinear map $T:V\times\dots\times V\to \mathbb{k}$ where we have $p$ copies of $V$.

Denote by $\Theta_p(V)$ the linear space of all tensors of a type $(p,0)$.

Definition: Let $T\in \Theta_p(V)$ and we say that $T$ is skew-symmetric tensor if for any $\sigma \in \Sigma_p$ we have $$(\sigma T)(v_1,\dots,v_p):=T(v_{\sigma(1)},\dots,v_{\sigma(p)})=(-1)^{\sigma}T(v_1,\dots,v_p),$$ where by $\Sigma_p$ I mean the group of all permutation of $p$ elements.

Let $\{e_1,\dots,e_n\}$ be a basis of $V$ then numbers $T_{i_1,\dots,i_p}:=T(e_{i_1},\dots,e_{i_p})$ are called coordinates of tensor $T$.

In coordinates the definition of skew-symmetric tensors can be written in this way:

$$(\sigma T)_{i_1,\dots,i_p}=T_{{i_{\sigma(1)},\dots,i_{\sigma(p)}}}.$$

And this is not so clear to me. Actually I have gotten the following:

Indeed, $(\sigma T)_{i_1,\dots,i_p}=(\sigma T)(e_{i_1},\dots,e_{i_p})=T(e_{\sigma(i_1)},\dots,e_{\sigma(i_p)})=T_{\sigma(i_1),\dots, \sigma(i_p)}$

And you see that this is completely different from above.

Can anyone explain to me what I am doing wrong, please?

Would be thankful for the detailed answer!

Answer 1

Let's apply the definitions a bit more carefully. Define $v_1,\dots,v_p$ to be the vectors $e_{i_1},\dots,e_{i_p}$. In other words: for $1 \leq k \leq p$, $v_k = e_{i_k}$. We then have $$ \begin{align} (\sigma T)_{i_1,\dots,i_p}&=(\sigma T)(e_{i_1},\dots,e_{i_p}) = (\sigma T)(v_1,\dots,v_p) = T(v_{\sigma(1)},\dots,v_{\sigma(p)}) \\ & = T(e_{i_{\sigma(1)}},\dots,e_{i_{\sigma(p)}}) = T_{i_{\sigma(1)},\dots,i_{\sigma(p)}}. \end{align} $$

Answer 2

I believe it is a matter of definitions. Start from here, where you are stuck,

$$(\sigma T)_{i_1,\dots,i_p}=T_{{\sigma(i_1),\dots,\sigma(i_p)}}\,\,.$$

Then consider the permutation $\sigma:\{1,\dots,p\}\to\{1,\dots,p\}$ and a subset $S=\{i_1,\dots,i_p\}\subseteq\{1,\dots,n\}$. In $S$ there are $p$ elements and you can permute them by using any permutation $\sigma\in\Sigma_p$, right? How do you define $\sigma(i_j)$, where $j\in\{1,\dots,p\}$? The natural thing is to define $$\sigma(i_j):= i_{\sigma_j}.$$ Notice that the group $\Sigma_p$ acts transitively on $S$ and actually on any subset of $p$ elements of $\{1,\dots,n\}$. Therefore,

$$(\sigma T)_{i_1,\dots,i_p}=T_{{\sigma(i_1),\dots,\sigma(i_p)}}=T_{{i_{\sigma(1)},\dots,i_{\sigma(p)}}}.$$

Notice also that, as $\sigma\in\Sigma_p$ and $i_j\in\{1,\dots,n\}$, there is no - a priori - a clear meaning of the writing $\sigma(i_j)$. However $j\in\{1,\dots,p\}$ and hence it has sense to say $\sigma(i_j):= i_{\sigma_j}.$

Edit

$\color{blue}{\text{Let's start with a bilinear map, this is the easiest case to deal with.}\\ \text{I used for a while a more common notation and then we do a comparison using the notation above.}\\ \text{I hope this may be more revealing for you.}}$

Let $T: V\times V\longrightarrow \Bbb K$, where $V$ is a vector space of dimension $n$. Fix a basis $\{e_1,\dots,e_n\}$. We write $T(e_j,e_k)=T_{jk}$ according to your notation above. $T$ is skew-symmetric iff the associated matrix is skew-symmetric that is iff $T_{jk}=-T_{kj}$. Indeed, for $\sigma\in\Sigma_2$ we have $$(\sigma\,T)(e_j,e_k)=T(e_{\sigma(j)},e_{\sigma(k)})=(-1)^{\sigma}T(e_j,e_k)$$ according to the definition above. Here $\sigma$ is acting on the set $\{j,k\}$.

$\color{blue}{\text{We now change the notation.}}$ So take two vectors of the basis above $e_{i_1}$ and $e_{i_2}$ and we do the same as above. Now we write $T(e_{i_1},e_{i_2})=T_{i_1i_2}$. According to the definition above $\sigma\in\Sigma_2$ we have $$(\sigma\,T)(e_{i_1},e_{i_2})=T(e_{\sigma(i_1)},e_{\sigma(i_2)})=(-1)^{\sigma}T(e_{i_1},e_{i_2})$$

If you think $i_1$ as $j$ and $i_2$ as $k$ then you get the description above, right? Here, the permutation $\sigma$ acts on $\{i_1,i_2\}$. Suppose $\sigma=(1\,\,2)$. Then $\sigma(i_1)=i_2$ and $\sigma(i_2)=i_1$, isn't it? So $\sigma$ is acting exchanging the sub-scriptures of $i_\bullet$. In other words, instead of looking at $\sigma$ as a map $\{i_1,i_2\}\to\{i_1,i_2\}$ you may see $\sigma$ as a map $\{1\,\,2\}\to\{1\,\,2\}$. Then we can say $$\sigma(i_j):= i_{\sigma_j} \quad \text{with} \quad j=1,2.$$

Notice that I have never really used that $p=2$ (in your notation). Indeed, maybe with a bit of work, the same holds for any $p$. Therefore, instead of looking at $\sigma\in\Sigma_p$ as a map $\{i_1\,\dots,\,i_p\}\to \{i_1\,\dots,\,i_p\}$, you can see $\sigma$ as a map from $\{1,\dots,p\}$ to itself. Again, you can see $$\sigma(i_j):= i_{\sigma_j} \quad \text{with} \quad j=1,\dots,p.$$

When you apply $\sigma$ to the multivector $(e_{i_1},\dots,e_{i_p})$ (thought as a $n\times p$ matrix) you are permuting the columns. Hence $\sigma(e_{i_j})$ can be only one among $e_{i_1},\dots,e_{i_p}$. No other choices. So permuting the $e_{i_j}$'s means permuting the $i_j$'s that means permuting the sub-subscriptures.