How to reduce an order 3 tensor to an order 2 tensor? Are there any techniques to reduce an order 3 tensor to an order 2 tensor?
For example, I have an $m \times m \times p$ tensor and I want to reduce it to a $m \times m \times 1$ tensor.
Thanks
 A: You multiply it with a vector of length $p$ summing over the $p$ components. The tensors I know have the same size in all indices.
A: I will give an example.
Let $T_{\mu\nu\rho}$ denote any tensor of order 3 (or, equivalently a tensor of type $(0,3)$).
The object
$$R_{\nu\rho}:=T_{\mu\nu\rho}A_{\mu},$$
where $A_{\mu}$ is any tensor of order 1 (or type $(0,1)$), is a tensor of order $2$ (or type $(0,2)$). This is the general contruction Ross Millikan refers to in his answer. We use the Einstein summation convention, whenever we repeat indices.
Formally speaking, you should prove that $R_{\nu\rho}$ transforms like a tensor of order $2$; in some old books a tensor is am mathematical object which transforms like a tensor, after all.
To do so, you need to consider the following formulae
$$T_{\mu\nu\rho}=T'_{abc}B_{a\mu}B_{b\nu}B_{c\rho},$$
$$A_{\mu}=A'_{\xi}B_{\xi\mu}, $$
which give you the transformation of the given tensors under a change of variable with orthogonal matrix $B_{\bullet\bullet}$, arriving at
$$R_{\nu\rho}:=T'_{abc}B_{a\mu}B_{b\nu}B_{c\rho}A'_{\xi}B_{\xi\mu}=(\text{we are dealing with scalar quantities})=B_{a\mu}(B_{\mu\xi})^tB_{b\nu}B_{c\rho}T'_{abc}A'_{\xi}=(\text{orthogonal change of variables})=\delta_{a\xi}B_{b\nu}B_{c\rho}T'_{abc}A'_{\xi}=
B_{b\nu}B_{c\rho}T'_{abc}A'_{a}=B_{b\nu}B_{c\rho}R'_{bc},$$
as claimed.
