Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

According to Wikipedia's page on tensor contraction:

In general, a tensor of type $(m,n)$ (with $m \geq 1$ and $n \geq 1$) is an element of the vector space $V \otimes \ldots \otimes V \otimes V^* \otimes \ldots \otimes V^*$ (where there are $m$ $V$ factors and $n$ $V^*$ factors). Applying the natural pairing to the $k$th $V$ factor and the $l$th $V^*$ factor, and using the identity on all other factors, defines the $(k,l)$ contraction operation, which is a linear map which yields a tensor of type $(m-1, n-1)$.

I must admit, I'm having trouble understanding this definition. What is the actual map? From what I gather, the $(k,l)$ contraction of, say, $$T = X_1 \otimes \ldots \otimes X_m \otimes \omega^1 \otimes \ldots \otimes \omega^n$$ is $$C(T) = \omega^l(X_k)\cdot X_1 \otimes \ldots \otimes \widehat{X_k} \otimes \ldots \otimes X_m \otimes \omega^1 \otimes \ldots \otimes \widehat{\omega^l} \otimes \ldots \otimes \omega^n,$$ where the hats indicate omission, and $C(T)$ is just my notation for contraction.

Is this correct? If so, is this often taken as the definition or are there other standard (equivalent) definitions?

share|cite|improve this question

Yes, this is correct, but I don't think you should get too used to contracting indices that aren't adjacent. Most of this formalism is quite general and in particular there are contexts where you have tensor products and duals but either there is no symmetry $X \otimes Y \to Y \otimes X$ or the symmetry is nontrivial in some way, and when you contract indices that aren't adjacent in this context what you are really doing is permuting some indices, contracting some indices, then permuting some more indices, and in general you have to keep track of what you're permuting.

To be more precise, in a braided monoidal category you have distinguished isomorphisms $\gamma_{A, B} : A \otimes B \to B \otimes A$ but you aren't guaranteed that $\gamma_{A, B} \gamma_{B, A} = \text{id}$, and even when you do have this condition $\gamma_{A, B}$ doesn't necessarily do the obvious thing. For example, on the category of $\mathbb{Z}$-graded vector spaces you can introduce the braiding $a \otimes b \mapsto (-1)^{|a| |b|} b \otimes a$ where $|a|, |b|$ refers to the grading. If you don't keep track of what permutations you're doing before and after you contract, you'll drop a sign.

share|cite|improve this answer
Thanks for the answer, though it might be a little while before I know enough category theory to fully appreciate it. I should have mentioned, though, that my motivation for asking was for the sake of doing explicit (but coordinate-free) computations in differential geometry. – Jesse Madnick Jun 23 '11 at 7:20
I love your category-based deontology :) – Mariano Suárez-Alvarez Jun 23 '11 at 18:04
@Mariano: well, one of the categories I'm talking about is the category of chain complexes. Surely you would agree that dropping signs in homological algebra is annoying and you should cultivate habits of thought that prevent you from doing it. – Qiaochu Yuan Jun 23 '11 at 20:21

Yes, that looks correct to me. I'm afraid I can't speak to what common alternative definitions are, however.

share|cite|improve this answer
Oh, hi, Zev. Mmm, okay, so one definition would be the calculation above and then extended by linearity? I guess that works. (I remember you emphasizing that elements of a tensor product are sums of things of the form $v \otimes w$.) – Jesse Madnick May 30 '11 at 5:21
Yes, that should be the definition of the contraction of an arbitrary tensor. – Zev Chonoles May 30 '11 at 5:26
@Jesse: you don't need to define anything elementwise. There is already a distinguished evaluation map $V \otimes V^{\ast} \to k$ and you are just tensoring this map with various identities and pre- and post-composing it with permutations of the factors of the tensor product. – Qiaochu Yuan May 30 '11 at 11:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.