On page 22 of Sean Carroll's Spacetime and Geometry, he says that tensors can act on other tensors and gives the following example:

$$ U^{\mu}_{\nu} = T^{\mu \rho}_{\sigma} S^{\sigma}_{\rho \nu}$$

where $T$ is a (2,1) tensor, $S$ is a (1,2) tensor, and $U$ is a (1,1) tensor.

I was trying to understand the derivation of this in terms of the tensor basis form: $$ T = T^{\mu \rho}_{\sigma} \hat{e}_{\mu}\otimes \hat{e}_{\rho}\otimes \hat{\theta}^{\sigma}, \; \; S = S^{\sigma}_{\rho \nu} \; \hat{e}_{\sigma}\otimes \hat{\theta}^{\rho} \otimes \hat{\theta}^{\nu} $$ where $\{ \hat{e}_{\mu} \}$ is the basis for the vector space and $\{ \hat{\theta}^{\mu} \}$ is the basis for the dual vector space.

Then, $TS = T^{\mu \rho}_{\sigma} S^{\sigma}_{\rho \nu} \; (\hat{e}_{\mu}\otimes \hat{e}_{\rho}\otimes \hat{\theta}^{\sigma}) (\hat{e}_{\sigma}\otimes \hat{\theta}^{\rho} \otimes \hat{\theta}^{\nu}).$ But I'm not sure how to proceed from here.

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    $\begingroup$ I doubt tensor product signs are necessary. Either way, Einstein summation notation works for both components and basis vectors among themselves, i.e. $a_\sigma \hat\theta^\sigma b^\sigma \hat e_\sigma$ is a scalar, just like $a_\sigma b^\sigma$ is when we disregarded the bases. Since our basis is orthonormal, $\hat\theta^\sigma \hat e_\sigma \implies \sum_\sigma \hat\theta^\sigma \hat e_\sigma = \sum_\sigma \delta_{\sigma\sigma} = 1$. $\endgroup$
    – acarturk
    Oct 1, 2019 at 10:12
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    $\begingroup$ But you took the Cartesian product of the tensors - not the tensor product .The tensor product would be $T \otimes S = T^{\mu \rho}_{\sigma} S^{\sigma}_{\rho \nu} \; \hat{e}_{\mu}\otimes \hat{e}_{\rho}\otimes \hat{\theta}^{\sigma} \otimes \hat{e}_{\sigma}\otimes \hat{\theta}^{\rho} \otimes \hat{\theta}^{\nu}$. Then apply acarturk's comment. $\endgroup$ Oct 2, 2019 at 5:58
  • $\begingroup$ And needless to say, $U = U^{\mu}_{\;\;\nu}\hat{e}_{\mu}\otimes \hat{\theta}^{\nu}$. $\endgroup$ Oct 2, 2019 at 23:07
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    $\begingroup$ I think that you should think about it as a tensor product that will produce a (3, 3)-Tensor, then apply contraction twice, as far as I understand this is the way that (acting tensor on a tensor is defined), and this logic will be similar to the one you are familiar with of acting a covector on a vector, you can form (1, 1)-tensor then contract to get (0, 0) tensor (or a real number), I'm not sure if this still a question you need an answer to, and I am relatively new to the topic, so if I made any mistake please correct me. $\endgroup$ Sep 7, 2022 at 5:51
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    $\begingroup$ Being strictly accurate about terminology can also help. A product involving tensors is not necessarily a "tensor product"! It might be a "scalar product" (also called "inner product") or an "outer product". In this example there are two parts involving inner product and one involving outer product. If in doubt you can always write each tensor as a sum of outer products of vectors or one-forms. $\endgroup$ May 18, 2023 at 16:55

1 Answer 1


I don't think there is an easy way to do this. Assuming that $U$ is the underlying vector space and $U^*$ is it's dual, you are staring with

$\mathbf{T} \in V=U\otimes U\otimes U^*$


$\mathbf{S} \in W=U\otimes U^*\otimes U^*$

and then seeking to define a unique linear map:

$\alpha: V\times W\to Q,\quad Q=U \times U^*$

The problem, I think, is that there is no unique way to define such a map. There are many ways to do it. So you will have to get into the specific indices.

Having said that, there is an easy way to understand the contraction of a vector ($A^\alpha\hat{e}_\alpha \in U$) and a 1-form ($S_\beta \hat{\theta}^\beta \in U^*$). By construction, $\hat{\theta}^\mu$ is a linear functional over $U$, i.e. $\hat{\theta}:U\to\mathbb{R}$ (or complex numbers, or integers etc). So what you are doing when contracting a vector and a 1-form, is to apply functional to the vector:

$\mathbf{S}\left(\mathbf{A}\right)=A^\alpha S_\beta \,\,\, \hat{\theta}^\beta\left(\hat{e}_\alpha\right)$

Appologies if I am spouting trivial things


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