let $S,T$ be respectively $k$-, $n$-tensors; $k,n>0$. Then we define the tensor product $$ T \otimes S(x_1,x_2,\ldots,x_{k+n}) := T(x_1,\ldots,x_k) S(x_{k+1}, \ldots, x_{k+n}) $$ (their product as real numbers, or product in any base field).

Now, what if either of $S,T$ is a $0$-tensor, i.e., a vector. How do we then define the product? I can tell the result would be a vector, i.e., $(0+0)$-tensor. Maybe we can use some duality to identify a $0$-tensor with a linear map, or we can use the inverse of the trace map? SERENITY NOW!!! (er, I mean, thanks for any help!).


$0$-tensors are just scalars, so the tensor product in this case is just scalar multiplication.

  • $\begingroup$ Thanks, that makes sense, but if 0-tensors are constants, what kind o tensors are vectors? $\endgroup$ – Gary. Jun 26 '15 at 23:00
  • $\begingroup$ They're $1-$tensors. $\endgroup$ – Glitch Jun 26 '15 at 23:01
  • $\begingroup$ But1-tensors are linear , i.e., 1-linear maps ( defined either in arguments in either $V$ or in $V^{*}$. Do we dualize someway to represent the linear maps as vectors and viceversa? $\endgroup$ – Gary. Jun 26 '15 at 23:03
  • $\begingroup$ Okay, I was cheating a bit by saying that a vector is a $1-$linear form. In fact the $1-$tensors are just linear forms, i.e. $1-$forms. These are canonically identified with vectors if we have an inner-product, though. $\endgroup$ – Glitch Jun 26 '15 at 23:07
  • 1
    $\begingroup$ To be clear, there are two types of "one tensor", element in $V$ is called $(1, 0)$-tensor, elemens in $V^*$ are called $(0,1)$-tensor. $\endgroup$ – user99914 Jun 26 '15 at 23:18

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