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A $(p,q)$-tensor can be defined as a multilinear function in several ways, which are mostly equivalent:

$$T:(V^*)^p\times V^q\to\mathbb R$$

or

$$T:V^q\to V^p$$

or

$$T:V^q\to \mathbb R\times V^p$$


The first definition seems to be the most common, but I want to define $V^*$ in terms of tensors, not vice-versa.


The second is what I want to use. In the case $q=0$, it gives a $p$-vector as a constant tuple of vectors

$$\{0\}\to V^p$$

and in the case $p=q=1$ it gives an ordinary linear transformation $V\to V$.

But in the case $p=0$, all $q$-covectors would be the same constant function

$$V^q\to\{0\}$$

so that doesn't seem to work.


The third would produce $q$-covectors correctly, as functions from $V^q$ to $\mathbb R$.

But it gives a $(1,0)$-tensor as

$$\{0\}\to\mathbb R\times V$$

which is a constant $(r,\vec v)$. Is this a problem? We can make any vector from these $(1,0)$-tensors by simply picking out $\vec v$ and ignoring $r$, or by multiplying $r\vec v$. (These are equivalent in the case $r=1$.) Of course, this correspondence is not one-to-one.


Any suggestions for a fourth definition, consistent with the question title?

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    $\begingroup$ The cases $p=0$ and $q=0$ require a convention which you seem to have misunderstood. For example, a $(0,q)$ tensor is a multilinear map $T:V^q\rightarrow \Bbb{R}$. The equivalencies you're using don't hold when $p=0$ or $q=0$. $\endgroup$ – Oliver Jones Aug 10 '18 at 7:59
  • $\begingroup$ What you're writing makes no sense; that's what I'm saying. $\endgroup$ – Oliver Jones Aug 10 '18 at 8:01
  • $\begingroup$ Your last two equivalencies contradict each other and so clearly you've misunderstood something. $\endgroup$ – Oliver Jones Aug 10 '18 at 8:03
  • $\begingroup$ What you're asking is how to express a mulitlinear map $T:(V^*)^p\rightarrow \Bbb{R}$ without using duals. I don't think this can be done. $\endgroup$ – Oliver Jones Aug 10 '18 at 8:16
  • $\begingroup$ And that is what? $\endgroup$ – Oliver Jones Aug 10 '18 at 8:21

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