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.

  • $\begingroup$ 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 '19 at 10:12
  • $\begingroup$ 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. $\endgroup$ – Cinaed Simson Oct 2 '19 at 5:58
  • $\begingroup$ And needless to say, $U = U^{\mu}_{\;\;\nu}\hat{e}_{\mu}\otimes \hat{\theta}^{\nu}$. $\endgroup$ – Cinaed Simson Oct 2 '19 at 23:07

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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