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.
 A: 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
