Is there any meaning behind tensor contraction. Or is it just randomly getting rid of some components by only selecting those with same index and sum them up?

For example, I know tensor is interpreted as a multilinear map. Maybe the contraction is doing some transformation to the map?

Or since matrix multiplication is also a tensor contraction. If we know what the meaning of matrix multiplication is and generalise it, we will know what tensor contraction means in a more general setting?

  • $\begingroup$ It’s a way of constructing a lower-rank tensor. $\endgroup$ – G. Smith Jul 5 at 0:02
  • $\begingroup$ Without contraction, there would be no way for tensors of different ranks to be related to one another. $\endgroup$ – G. Smith Jul 5 at 0:15
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    $\begingroup$ Start with a matrix and intuit its trace, geometrically, if you were so inclined. $\endgroup$ – Cosmas Zachos Jul 5 at 0:38
  • $\begingroup$ Suppose you summed over three indices. Or two covariant ones. Or suppose the sum didn’t include all of the components. Then you would not get another tensor. So contraction is defined the way it is because it is the only way to get a lower-rank tensor from a higher-rank one. $\endgroup$ – G. Smith Jul 5 at 0:47

I don't know if you'll find the following helpful, but let $V$ be a vector space and let $T^i_j$ be a $\binom11$-tensor.

Probably you are aware that $V^\ast\otimes V \cong \operatorname{End}(V) \cong \mathcal M_n(\mathbb R)$, the space of $n\times n$ real matrices, where $n$ is the dimension of $V$. If you are not, then just convince yourself that the element $\phi\otimes v\in V^\ast\otimes V$ gives you an endomorphism of $V$ by $x\mapsto \phi(x)v$, and that extending this map by linearity we get an isomorphism.

The contraction $T^i_i$ is simply the trace of this matrix, as you probably also know. For some geometric interpretations of the trace, see this mathoverflow post.

Now let's consider a general $\binom mn$-tensor $T^{i_1\ldots i_m}_{j_1\ldots j_n}$. Let's say for ease of notation that we want to contract $i_1$ with $j_1$. We can view $T^{i_1\ldots i_m}_{j_1\ldots j_n}$ as a matrix $T^{i_1}_{j_1}$ whose entries are $\binom {m-1}{n-1}$-tensors, namely $T^{i_2\ldots i_m}_{j_2\ldots j_n}$. The contraction of $j_1$ with $j_2$ is the trace of this matrix, which is a $\binom {m-1}{n-1}$-tensor as well.


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