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In physics one would speak of a tensor of second rank having nine components (in three dimensions) usually written as

$$T = \begin{bmatrix} t_{11} & t_{12} & t_{13} \\ t_{21} & t_{22} & t_{23} \\ t_{31} & t_{32} & t_{33}\end{bmatrix}$$ and speaks of how it is changes in other coordinate systems.

Mathematics speaks of tensors from tensor products such as $V\otimes V$ and $V^* \otimes V^*$ where for example $V=\mathbb{R}^3$.

Now my confusion that be illustrated by the following example of a physics type tensor which is the direct product

$$T = U V^t= \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3}\end{bmatrix} \begin{bmatrix} v_{1} & v_{2} & v_{3} \end{bmatrix}$$

Is this physics tensor suppose to be identified with a mathematics style tensor in $\mathbb{R}^3 \otimes \mathbb{R}^3$ or in ${\mathbb{R}^3}^* \otimes {\mathbb{R}^3}^*$?

Identified with a mathematics style tensor $U\otimes V \in \mathbb{R}^3 \otimes \mathbb{R}^3$, then if the coordinate transform that the physics definition speaks of is written as $U'=AU$ and $V'=AV$ where $$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{bmatrix}$$ the way the physics tensor transform rule comes about from $$T' = U'{V'}^t=AUV^tA^t=ATA^t$$

But since the main notion of a tensor is suppose to be a multilinear form then aren't we suppose to think of it as in ${\mathbb{R}^3}^* \otimes {\mathbb{R}^3}^*$? So shouldn't we identify it with a mathematics style tensor $U\otimes V \in {\mathbb{R}^3}^* \otimes {\mathbb{R}^3}^*$ because we can think of the function $f(x,y) = y^tTx = y^tUV^tx$ constructed from the linear functionals $y^tU$ and $V^tx$. If the coordinate transform that the physics definition speaks of is written as $x'=Ax$ and $y'=Ay$ then the way the physics tensor transform rule comes about from ${y'}^tT'x' = y^tA^tT'Ax = y^tTx$

Which is the right way to identify the physics tensor with the mathematics view and what is the right way for the the way the physics tensor transform rule to come about?

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  • $\begingroup$ It seems that it depends on whether $U, V \in \mathbb R^3$ or ${\mathbb R^3}^*$. $\endgroup$ – user99914 Dec 2 '14 at 10:32
  • $\begingroup$ take a look to en.wikipedia.org/wiki/Dyadics $\endgroup$ – janmarqz Dec 3 '14 at 2:12
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    $\begingroup$ @janmarqz: is Wikipedia credible? ☺ By the way, I think that explanations similar to your answer can be found in textbooks on tensor calculus. These are basics. $\endgroup$ – Incnis Mrsi Dec 10 '14 at 21:10
  • $\begingroup$ @IncnisMrsi: Try to question me. And, besides wikipedia (in english) is massively scanned and improved. Several friend around will confirm my claims. $\endgroup$ – janmarqz Dec 10 '14 at 21:15
  • $\begingroup$ people's confusion comes from the blurring understanding of duality, not distinguishing for a n-tuple of data from row or columns matrices makes the life a suffering ;) $\endgroup$ – janmarqz Dec 10 '14 at 21:33
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Ok, in 3-dimensional real vector space, $V$, each tensor of rank two can be considered as a bilinear for in each of the following 4 cases:

If $B\in V^*\otimes V^*$ then $B$ is a pairing $V\times V\to\Bbb{R}$
via $(v,w)\to B(v,w)=v^{\top}Bw$ or in components $B(v,w)=v^sw^tB_{st}$;

If $B\in V\otimes V$ then $B$ is a pairing $V^*\times V^*\to\Bbb{R}$
via $(f,g)\to B(f,g)=fBg^{\top}$ or in components $B(f,g)=f_sg_tB^{st}$;

If $B\in V^*\otimes V$ then $B$ is a pairing $V\times V^*\to\Bbb{R}$
via $(v,f)\to B(v,f)=v^{\top}Bf^{\top}$ or in components $B(v,f)=v^sf_t{B_s}^t$;

If $B\in V\otimes V^*$ then $B$ is a pairing $V^*\times V\to\Bbb{R}$
via $(f,v)\to B(f,v)=fBv$ or in components $B(f,v)=f_sv^t{B^s}_t$.

Where $v=v^se_s=\left(\begin{array}{c} v^1\\ v^2\\ v^3 \end{array}\right)$ for a vector (similar for $w$), and $f=f_s\varepsilon^s=(f_1,f_2,f_3)$ for a covector (similar for $g$).

You can see how the 4 ways to index with two indexes are used efficiently.

At the time when there is a change of basis one uses $b_k={C^s}_ke_s$ and $\beta^k={A_s}^k\varepsilon^s$ to find how the tensor $B$ changes in each of the 4 cases above.

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