# Possible error in Wikipedia article for $(p,q)$ tensor space notation

I am currently studying tensor space theory (over a field) and was reading through the generic Wikipedia article for tensors when I cam across the following:

A type $$(p,q)$$ tensor $$T$$ is a multilinear map, $$T: \underbrace{V^*\times...\times V^*}_\text{p-copies}\times\underbrace{V\times...\times V}_\text{q-copies} \to K$$ where $$K$$ is the underlying field.

Where the author(s) then goes onto define the notation for the tensor product as

$$\underbrace{V\otimes...\otimes V}_\text{p-copies} \otimes \underbrace {V^* \otimes...\otimes V^*}_\text{q-copies}$$

My question: why did they flip the usage of the $$p$$'s and $$q$$'s? Namely, they first called copies of the dual space $$p$$ and copies of the vector space $$q$$ $$-$$ then went on to reverse this in the notation for the tensor product (calling copies of the vector space $$p$$ and copies of the dual $$q$$). Is this an error that should be corrected? I also noticed this in a lecture by Frederic P. Schuller on tensor space theory. He did the exact same thing when defining what a tensor is and then the tensor product (at minute $$29:23$$).

Please correct me if I am wrong, as I am new to differential geometry and tensors.

• Let's start with what a $(0,1)$-tensor $T$ is. By the definition given, it is a linear function $T: V \rightarrow K$ and therefore is an element in the dual vector space $V^*$. On the other hand, a $(1,0)$-tensor is a linear function $T: V^* \rightarrow K$ and therefore is an element in $(V^*)^* = V$. Etc. Feb 15, 2021 at 21:30
• @Deane : pedantically crossing tees: assuming that our vector spaces are finite dimensional... as they are meant (of course) to be here and in Wikipedia, one supposes. Feb 15, 2021 at 21:34
• @Deane - ahhh, I see what you mean. Thank you very much for clarifying! I think I have been staring at this stuff for too long - probably time for a coffee break! Feb 15, 2021 at 21:35

Note that a $$(0,1)$$-tensor is a linear map $$V \to K$$, so $$V^*$$ is the space of such tensors. Similarly, a $$(1,0)$$-tensor is a linear map $$V^* \to K$$, so $$V^{**}$$ (which can be canonically identified with $$V$$ if we are in finite dimensions) is the space of such tensors.
Now, if $$v \in V$$ and $$f \in V^*$$, define $$v \otimes f : V^* \times V \to K$$ by $$(v \otimes f)(g,w) = g(v)f(w)$$. Note then that $$v \otimes f \in V \otimes V^*$$, not $$v \otimes f \in V^* \otimes V$$. Thus $$v \in V,\, f \in V^* \implies v \otimes f \in V \otimes V^*$$ as naturally one would like it to happen.
No, this is correct. The asterisk denotes the dual, i.e. if $$V$$ is a $$K$$-vector space, $$V^*$$ is the vector space of linear maps $$V \to K$$. Intuitively, in the first case, you have something very similar (except with multilinear maps). In the second case, you are missing this step (the “$$\to K$$” part), so you have to dualize everything in the domain, which gives you the swapped asterisks (as $$V^{**} = V$$ for finite-dimensional vector spaces).