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I have a question regarding tensors and basis changes, however upon searching the web I've found an infinity of definitions for tensors so I'll have to give the one I know first:

Given a vector space $E$ over a (commutative) field $F$, we call, for $p\ge 1$ a $p$-covariant tensor a map of the type $$ f:E\times\overset{\text{(p)}}{\dotsb}\times E\to F $$ where $f$ is $p$-linear (linear in each of the $p$ components). We call the set of $p$-covariant tensors (over $E$) $ T_p(E)$, which is the set of $p$-linear maps from $E$ to $F$, in other words, $T_p(E) = \mathcal L_p(E,F)$.

Next we define for $f \in T_p(E),\ g \in T_q(E)$ the tensor product of $f$ and $g$, as $f \otimes g \in T_{p+q}(E)$, where: $${f\otimes g : E\times\overset{\text{(p)}}{\dotsb}\times E\times\overset{\text{(q)}}{\dotsb}\times E \to F} \atop {(u_1,\dots,u_p,v_1\dots,v_q) \mapsto f(u_1,\dots,u_p)\cdot g(v_1\dots,v_q)} $$

For example, for $p=1, \ T_1(E) = E^*$. For arbitrary $p$, it can be shown that $T_p(E)$ is a vector space over $F$, and if $e_1,\dots,e_n$ is a basis for $E$, with dual basis $e_1^*,\dots,e_n^*$ ($e_i^*(e_j) = \delta_{ij}$) then a basis for $T_p(E)$ is $\{e_{i_1}^*\otimes\dots\otimes e_{i_p}^*\}_{1\le i_1,\dots, i_p\le n}$

My question regards 2-covariant tensors. For simplicity, say we have $\varphi\in T_2(E)$, and $E$ is two-dimensional with basis $e_1,e_2$. Then $T_2(E)$ has the ordered basis $B_e = \{e_1^*\otimes e_1^*, \ e_1^*\otimes e_2^*, \ e_2^*\otimes e_1^*, \ e_2^*\otimes e_2^*\}$ and $\varphi = (a,b,c,d)$ in this basis.

What I want to know is, if $B_u$ is a similar basis of $T_2(E)$ relative to basis $u_1, u_2$ of $E$, how can I express $\varphi$ in basis $B_u$, using the fact that I know the basis change matrix that changes from $e_1,e_2$ to $u_1,u_2$ in $E$? I know it has to do with a Kronecker product of matrices, but I don't quite understand how it works.

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1 Answer 1

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Let $B$ and $B'$ be two bases of $E$ and denote by $A$ the matrix of base change from $B$ to $B'$. First there are the induced dual bases on $E^\ast = T_1(E)$, denoted by $B^\ast$ and $B'^\ast$. The matrix of base change from $B^\ast$ to $B'^\ast$ is $A^\ast:= {}^t A^{-1}$ as you can easily check.

Now you have induced bases $\tilde{B}$ and $\widetilde{B'}$ on $E^\ast \otimes E^\ast := T_2(E)$ with your notations. The matrix of base change from $\tilde{B}$ to $\widetilde{B'}$ is the Kronecker product of $A^\ast$ with itself.

Look at the Wikipedia page if you don't know about Kronecker products.

I think that the best way to understand this stuff is to learn tensor products of vector spaces and not only tensor products of maps.

Edit: In order to prove these things, you have to write carefully the matrix of basis changes. Write $B = (e_1,\dots,e_n)$ and $B' = (f_1,\dots,f_n)$. By definition of $A$, $f_i = \sum_{j=1}^n A_i^j e_j$ (you can also adopt the other convention). Now you want the matrix from $B^\ast$ to $B'^\ast$. This matrix $C$ should satisfy $f_i^\ast = \sum_{j=1}^n C_i^j e_j^\ast$. Evaluating at $f_k$ you find, $\delta_i^k = \sum_{j=1}^n C_i^j e_j^\ast(f_k)$. But $f_k$ is $\sum_{l=1}^n A_k^l e_l$. So $\delta_i^k = \sum_{j=1}^n C_i^j A^j_k$. These equations are satisfied for $C = {}^t A^{-1}$ which shows the first part.

For the tensor product, you have to understand how to write $f_{j_1}^\ast \otimes f_{j_2}^\ast$ with $e_{i_1}^\ast \otimes e_{i_2}^\ast$. We have $f_{j_1}^\ast \otimes f_{j_2}^\ast = (\sum_{i_1} C_{j_1}^{i_1} e_{i_1}^\ast) \otimes (\sum_{i_2} C_{j_2}^{i_2} e_{i_2}^\ast) = \sum_{i_1, i_2} C_{j_1}^{i_1} C_{j_2}^{i_2} e_{i_1}^\ast \otimes e_{i_2}^\ast$.

Compare these coefficients with the definition of the Kronecker product; you find the same thing (with a certain ordering of the bases in $T_2(E)$).

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  • $\begingroup$ Thanks, is a proof to be found somewhere? $\endgroup$
    – GPerez
    Commented Jan 3, 2014 at 18:38

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