I just read a proof showing that the stress tensor is a contravariant second order tensor and I cannot reconcile it with what I already knew of second order tensors.

A linear transformation $T\colon V\to V$ is a mixed second order tensor. A bilinear form $B \colon V \times V \to \mathbb{R}$ is a second order covariant tensor. Given that the stress tensor takes a vector and gives back a stress vector I would have expected it to be a mixed tensor like the linear transformation unless somehow its being thought of as a function $V\to V^*$ or $V^* \times V^*\to \mathbb{R}$.

  • 2
    $\begingroup$ I would say that the stress tensor takes a plane (i.e. a 1-form) and gives a vector. $\endgroup$
    – Javier
    Jun 6, 2018 at 0:16
  • $\begingroup$ there is an isomorphism ${\rm hom}(V,V)\cong V\otimes V^*$ $\endgroup$
    – janmarqz
    Jun 7, 2018 at 21:46
  • $\begingroup$ @janmarqz Ok that makes sense. I haven't seen an introduction to tensors explain stress tensors as that. $\endgroup$
    – user782220
    Jun 8, 2018 at 16:56
  • $\begingroup$ a rank two mixed tensor is bilinear map $V^*\times V\to\Bbb R$ and the set of all of these maps is the vector space $V\otimes V^*$. $\endgroup$
    – janmarqz
    Jun 8, 2018 at 18:18
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    $\begingroup$ @janmarqz I would give your answer the check mark if it wasn't a comment. $\endgroup$
    – user782220
    Jun 10, 2018 at 0:45

1 Answer 1


Let me reframe this in other words:

If $V$ is finite dimensional vector space over the field $\Bbb F$ then there is an isomorphism $\hom(V,V)\to V\otimes V^*$.

Here, $\hom(V,V)$ is the vector space of linear transformations $V\to V$, and $V\otimes V^*$ a the space of rank two mixed tensors which correspond to the set of all bilinear maps $V^*\times V\to\Bbb F$.

  • $\begingroup$ In the case of the stress tensor being a contravariant second order tensor would this be $\hom(V^*, V) \to V \otimes V$ $\endgroup$
    – user782220
    Jun 10, 2018 at 20:57
  • $\begingroup$ yes, these are the rank two contravariant tensor, but take into account their components are in this case, indexed above, as $T^{ij}$. $\endgroup$
    – janmarqz
    Jun 11, 2018 at 1:25
  • $\begingroup$ @user782220: if you already grasp the isomorphism device, would you like to see other is the line of thoughts at math.stackexchange.com/questions/2811697/… there, there're more details in terms of components $\endgroup$
    – janmarqz
    Jun 11, 2018 at 16:14

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