# How can the stress tensor be a contravariant second order tensor?

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 $V^* \times V^*\to \mathbb{R}$.

• I would say that the stress tensor takes a plane (i.e. a 1-form) and gives a vector. Jun 6, 2018 at 0:16
• there is an isomorphism ${\rm hom}(V,V)\cong V\otimes V^*$ Jun 7, 2018 at 21:46
• @janmarqz Ok that makes sense. I haven't seen an introduction to tensors explain stress tensors as that. Jun 8, 2018 at 16:56
• 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 $V\otimes V^*$. Jun 8, 2018 at 18:18
• @janmarqz I would give your answer the check mark if it wasn't a comment. Jun 10, 2018 at 0:45

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 $$V^*\times V\to\Bbb F$$.
• In the case of the stress tensor being a contravariant second order tensor would this be $\hom(V^*, V) \to V \otimes V$ Jun 10, 2018 at 20:57
• yes, these are the rank two contravariant tensor, but take into account their components are in this case, indexed above, as $T^{ij}$. Jun 11, 2018 at 1:25