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.

  • 3
    $\begingroup$ 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. $\endgroup$
    – Deane
    Feb 15, 2021 at 21:30
  • 2
    $\begingroup$ @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. $\endgroup$
    – peter a g
    Feb 15, 2021 at 21:34
  • $\begingroup$ @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! $\endgroup$ Feb 15, 2021 at 21:35

2 Answers 2


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).


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