# Computing Tensors, Alternating Tensors, and Wedge Product

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?

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

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 $$1/2$$.