# Tensor acting on another tensor

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.

• 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, $\hat\theta^\sigma \hat e_\sigma \implies \sum_\sigma \hat\theta^\sigma \hat e_\sigma = \sum_\sigma \delta_{\sigma\sigma} = 1$. – acarturk Oct 1 at 10:12
• 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. – Cinaed Simson Oct 2 at 5:58
• And needless to say, $U = U^{\mu}_{\;\;\nu}\hat{e}_{\mu}\otimes \hat{\theta}^{\nu}$. – Cinaed Simson Oct 2 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

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

and

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

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