Question: Given the tensors below over $\mathbb{R}^n$, we take the standard basis $e_1,\cdots,e_n$ for $\mathbb{R}^n$, and the corresponding standard dual basis, and the induced standard basis on all related tensor spaces. Compute the following:

  1. $(e^1\otimes e^2)(e_1\otimes e_2)$
  2. $(e^1\otimes e^2)(e_1\otimes e_3)$
  3. $\operatorname{Alt}(e^1\otimes e^2)(e_1\otimes e_2)$
  4. $\operatorname{Alt}(e^1\otimes e^2)(\operatorname{Alt}(e_1\otimes e_2))$
  5. $(e^1 \wedge e^2)(e_1\otimes e^2)$

Progress: I understand what tensors generally are, and the fact that $e^1$ represents a $(0,1)$-tensor as it takes in a vector as an input, and $e_1$ is a $(1,0)$-tensor as it takes a covector (dual vector) as an input. I'm stuck on how to apply this knowledge in calculating these tensor products.

For (1), the inputs are $e_1$ and $e_2$, and thus the tensor product can be expanded to represent $e^1(e_1)\cdot e^2(e_2)$ which should equal one. Does that mean (2) is zero.

I know that a $(0,k)$-tensor or $(k,0)$-tensor is alternating if swapping a pair of input should negate the inputs, but I am unaware of how to apply this in computation.

I also know that in terms of the standard basis the wedge product of $e^1\wedge e^2 = e^{12}-e^{21}$. But again, I am still unaware of how to apply this.

I am having trouble applying the concepts of what I learned to evaluate tensor products, any hints?

  • $\begingroup$ Shouldn't the last be $(e^1\wedge e^2)(e_1\otimes e_2)$ to make it scalar contraction? $\endgroup$ Commented Jun 13, 2022 at 9:54

1 Answer 1


You are correct about the first two claims. I'll do #4 since that one seems the most involved, and then you can probably do the others.

First, a point about notation. I interpret 'Alt' as an operator which takes in a tensor, and returns an alternating tensor. This can be defined in two ways. If we have a rank $k$ tensor, then we look at the symmetric group $S_k$ of self-bijections of $k$ symbols. Then each of these has a sign, given by the parity of the number of transpositions it takes to reach that transposition from the identity. That this is well-defined is a standard fact from group theory, which I'm sure you can find discussed on this site. Then we allow these bijections, which we denote by $\sigma \in S_k$ to act by permuting the tensor factors, and if $T$ is a tensor, we denote the tensor we obtain by applying this permutation by $T^\sigma$. We can define $Alt(T)$ by either $\sum_{\sigma \in S_k} (-1)^{(\operatorname{sgn}\sigma)}T^\sigma$, or by $\frac{1}{k!}\sum_{\sigma \in S_k} (-1)^{(\operatorname{sgn}\sigma)}T^\sigma$. The presence of the factor $1/k!$ should be thought of as a normalization factor, which is convenient in some settings, but ultimately not that important, so I'll ignore it for your question, which is just about a $2-$tensor anyway.

Now, what is Alt of $e_1 \otimes e_2$? Well there is only one non-trivial permutation, its sign is $-1$, and so we obtain $e_1 \otimes e_2 - e_2 \otimes e_1$. We get something similar for upper indices, and we know that tensors are multilinear, so we have $$(e^1 \otimes e^2 - e^2 \otimes e^1)(e_1 \otimes e_2 - e_2 \otimes e_1) = (1 - 0) - (0 - 1) = 1 + 1 = 2$$

Worth noting is that if your convention requires us to divide out by $2!= 2$ in both terms, then we'd have to divide this by $4$ and we'd only get $1/2$.


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