I've already known the definition of the tensor that a tensor T of type $(k,l)$ is a multilinear map from a collection of dual vectors and vectors to $\mathbf{R}$: $$ T: \underbrace{T^\ast_p\times\cdots\times{}T^\ast_p}_{k\text{ times}}{\,}\times{\,}\underbrace{T_p\times\cdots\times{}T_p}_{l\text{ times}}\to \mathbf{R}$$ $"\times"$denotes the Cartesian products and $T^\ast_p$ , $T_p$ is respectively cotangent space and tangent space.

A tensor $T$ can be written as $T=T^{\mu_1\cdots\mu_k}_{\phantom{\mu_1\cdots\mu_k}\nu_1\cdots\nu_l} \hat{e}_{(\mu_1)}\otimes\cdots\otimes\hat{e}_{(\mu_k)}\otimes\hat{\theta}^{(\nu_1)}\otimes\cdots\otimes\hat{\theta}^{(\nu_l)}$.

My problem is how to understand a tensor acts on a tensor to be another tensor explicitly as the multilinear maps. For example, $$U^\mu_{\phantom{\mu}\nu}=T^{\mu\rho}_{\phantom{\mu\rho}\sigma}S^{\sigma}_{\phantom{\sigma}\rho\nu}$$ explicitly in $$U^\mu_{\phantom{\mu}\nu}\hat{e}_{(\mu)}\otimes\hat{\theta}^{(\nu)}=T^{\mu\rho}_{\phantom{\mu\rho}\sigma}\hat{e}_{(\mu)}\otimes\hat{e}_{(\rho)}\otimes\hat{\theta}^{(\sigma)}S^{\sigma}_{\phantom{\sigma}\rho\nu}\hat{e}_{(\sigma)}\otimes\hat{\theta}^{(\rho)}\otimes\hat{\theta}^{(\nu)}$$ What map does $T^{\mu\rho}_{\phantom{\mu\rho}\sigma}S^{\sigma}_{\phantom{\sigma}\rho\nu}$ precisely mean and why does it equal to the map denoted by $U^\mu_{\phantom{\mu}\nu}$ ?


1 Answer 1


There are two relevant operations you want to be familiar with on tensors.

First, given a $(m,n)$ tensor $T$ and an $(k,l)$ tensor $S$, you can construct their tensor product $T \otimes S$ which will be a $(m + k, n + l)$ tensor. Invariantly,

$$ (T \otimes S)(\varphi^1, \ldots, \varphi^m, \varphi^{m+1}, \ldots, \varphi^{m + k}, v_1, \ldots, v_n, v_{n+1}, \ldots, v_{n+l}) := \\ T(\varphi^1, \ldots, \varphi^m, v_1, \ldots, v_n) \cdot S(\varphi^{m+1}, \ldots, \varphi^{m+k}, v_{n+1}, \ldots, v_{n+l}). $$

If you choose a basis $(e_1, \ldots, e_n)$ for $V$ and the corresponding dual basis $(e^1, \ldots, e^n)$ for $V^{*}$ then if

$$ T = T^{\mu_1 \ldots \mu_m}_{\phantom{\mu_1 \ldots \mu_m}\nu_1 \ldots \nu_n} e_{\mu_1} \otimes \ldots \otimes e_{\mu_m} \otimes e^{\nu_1} \otimes \ldots \otimes e^{\nu_n}, \\ S = S^{\rho_1 \ldots \rho_k}_{\phantom{\rho_1 \ldots \rho_k} \sigma_1 \ldots \sigma_l} e_{\rho_1} \otimes \ldots \otimes e_{\rho_k} \otimes e^{\sigma_1} \otimes \ldots \otimes e^{\sigma_l} $$

then $ T \otimes S $ will be represented by

$$ T^{\mu_1 \ldots \mu_m}_{\phantom{\mu_1 \ldots \mu_m}\nu_1 \ldots \nu_n} S^{\rho_1 \ldots \rho_k}_{\phantom{\rho_1 \ldots \rho_k} \sigma_1 \ldots \sigma_l} e_{\mu_1} \otimes \ldots \otimes e_{\mu_m} \otimes e_{\rho_1} \otimes \ldots \otimes e_{\rho_k} \otimes e^{\nu_1} \otimes \ldots \otimes e^{\nu_n} \otimes e^{\sigma_1} \otimes \ldots \otimes e^{\sigma_l}. $$

Second, given an $(m,n)$ tensor $T$ with $m, n \geq 1$ and a choice of indices $1 \leq k \leq m$ and $1 \leq l \leq n$, you can define a $(m-1,n-1)$ tensor called the contraction or trace of $T$ with respect to $(k,l)$. To define it, choose some basis $(e_1, \ldots, e_n)$ for $V$ and define

$$ \mathrm{tr}_{(k,l)}(T)(\varphi^1, \ldots, \varphi^{m-1}, v_1, \ldots, v_{n - 1}) := T(\varphi^1, \ldots, \varphi^{k-1}, e^\sigma, \varphi^{k}, \ldots, \varphi^{m-1}, v_1, \ldots, v_{l-1}, e_\sigma, v_{l}, \ldots, v_{n - 1}). $$

You can check that this definition does not depend on the basis $(e_1, \ldots, e_n)$. In particular, if you choose a basis $(e_1, \ldots, e_n)$ for $V$ and represent $T$ as

$$ T = T^{\mu_1 \ldots \mu_m}_{\phantom{\mu_1 \ldots \mu_m}\nu_1 \ldots \nu_n} e_{\mu_1} \otimes \ldots \otimes e_{\mu_m} \otimes e^{\nu_1} \otimes \ldots \otimes e^{\nu_n} $$

then $ \mathrm{tr}_{(k,l)}(T) $ will be represented as

$$ T^{\mu_1 \ldots \mu_{k-1} \sigma \mu_{k+1} \mu_{m}}_{\phantom{\mu_1 \ldots \mu_{k-1} \sigma \mu_{k+1} \mu_{m}}\nu_1 \ldots \nu_{l-1} \sigma \nu_{l+1} \ldots \nu_n} e_{\mu_1} \otimes \ldots e_{\mu_{k-1}} \otimes e_{\mu_{k+1}} \ldots \otimes e_{\mu_m} \otimes e^{\nu_1} \otimes \ldots \otimes e^{\nu_{l-1}} \otimes e^{\nu_{l+1}} \otimes \ldots \otimes e^{\nu_n}. $$

That is, the representation of $\mathrm{tr}_{(k,l)}(T)$ is obtained by choosing a covariant and a contravariant index and "canceling" (contracting) them out using summation.

The name trace is justified by the fact that if $T$ is a $(1,1)$ tensor then you can easily check that the $(1,1)$ trace of $T$ is the same as the trace of $T$ considered as a linear map $V \to V$ (or $V^{*} \to V^{*}$).

Returning to your question, $T$ is a $(2,1)$ tensor and $S$ is a $(1,2)$ tensor. The $(1,1)$ tensor $U$ is obtained by taking the tensor product $T \otimes S$ of $T$ and $S$ (resulting in a $(3,3)$ tensor) and then contracting twice. More explicitly,

$$ U = \mathrm{tr}_{(2,1)} (\mathrm{tr}_{(3,1)} (T \otimes S)) = \mathrm{tr}_{(2,1)} (\mathrm{tr}_{(2,2)} (T \otimes S) ).$$

  • $\begingroup$ You're welcome. The contraction of the product is a different object, not a shorter way of writing something. BTW, Many operations in linear algebra have this form of multiplication and then contraction. For example, if $T$ is a $(1,1)$-tensor and $v$ is a $(1,0)$ tensor, then the contraction of $T \otimes v$ is just $T(v)$ when $T$ is considered as a linear map and $v$ is considered as a vector in $V$. More generally, multiplication of matrices is an example of multiplying and then contracting. $\endgroup$
    – levap
    Nov 7, 2015 at 16:27
  • $\begingroup$ sorry for the wrong clicking on the delete button... I think I've roughly understood the contraction of the tensor. A new question, from the definition of the contraction, $T^\mu$ is the contraction of $T^{\mu\sigma}_{\phantom{\mu\sigma}\sigma}$ means that when $T^{\mu\sigma}_{\phantom{\mu\sigma}\sigma}$ acts on the $(\varphi^{\mu},e^{\sigma},e_{\sigma})$, the result equals to the result of $T^{\mu}$ acting on $(\varphi^{\mu})$, but what if $T^{\mu\sigma}_{\phantom{\mu\sigma}\sigma}$ acts on another $(\varphi^1,\varphi^2,\varphi_3)$, $\endgroup$
    – R. Pan
    Nov 7, 2015 at 16:49
  • $\begingroup$ does $T^{\mu}$ have any connection to the $T^{\mu\sigma}_{\phantom{\mu\sigma}\sigma}$ in this case? Maybe I did't understand the meaning of " this definition does not depend on the basis $(e_1,\cdots,e_n)$". $\endgroup$
    – R. Pan
    Nov 7, 2015 at 16:58
  • $\begingroup$ The tensor $T^{\mu \sigma}_{\phantom{\mu \sigma} \sigma}$ is a $(1,0)$ tensor so it cannot act on a triple, only on a covector. We have: $T^{\mu \sigma}_{\phantom{\mu \sigma} \sigma} (v_i e^i) = T^{i \sigma}_{\phantom{i \sigma} \sigma} v_i$ where we have here two summations both on $\sigma$ and $i$. $\endgroup$
    – levap
    Nov 7, 2015 at 17:13
  • $\begingroup$ Now from my understand, when I write $$U^{\mu}_{\phantom{\mu}\nu}=T^{\mu\rho}_{\phantom{\mu\rho}\sigma}S^{\sigma}_{{}{\,\rho\nu}}$$ it means I define a contraction of the result of the tensor product which is $T^{\mu\rho}_{\phantom{\mu\rho}\sigma}\otimes{}S^{\sigma}_{\phantom{\,}\rho\nu}$ and use $U^{\mu}_{\phantom{\mu}\nu}$ to represent the contraction. The symbol "$=$" just means the definition of $U^{\mu}_{\phantom{\mu}\nu}$, not commonly the meaning "equal to". Am I right? $\endgroup$
    – R. Pan
    Nov 8, 2015 at 13:00

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