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I have come across various sources that talk about traces of tensors. How does that work? In particular, there seem to be such an equality:

$$ \text{Tr}(T_1\otimes T_2)=\text{Tr}(T_1)\text{Tr}(T_2)\;\;\;...(1) $$

as found in this Stackexchange question and this Wikipedia page.

The Wiki page "Tensor Contraction" speaks of tensor contraction as some generalization of trace, though without providing any formulation or example.

My questions: How do they all work? What is trace for a tensor? How does such trace interact with tensor product?

In particular, I have this contraction: (following Einstein's summation convention) $$ F^{\mu\nu}F_{\mu\nu} $$ where $F$ is a rank-2 tensor and each $F_{\mu\nu}$ is a $4\times 4$ matrix. Can it be expressed as trace of some sort? Subsequently can I apply (1) to split the expression into product of, say, the trace of $F$?

Additionally, $F^{\mu\nu}$ is anti-symmetric and I am trying to prove the above equals to zero. So being able to use (1) can be awesome.

Note: I have some although limited background in differential geometry and algebra. English words are great. But please supply formal definitions as well. At the same time, explanations with as little abstract algebraic constructions as possible would be much appreciated. Focus on finite dimension is fine. Extension to separable Hilbert space, partial trace and etc is welcomed too.


EDIT: The second floor to in this post seems to be good. I don't understand, however, how tensor product of matrices work? Still, when does contraction come into play?

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    $\begingroup$ Possible duplicate of How to compute the trace of a tensor? $\endgroup$
    – Alex M.
    Commented Sep 18, 2018 at 17:05
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    $\begingroup$ Nope. Not even close. (If you are interested, kindly read the answer below? I haven't gotten a chance.) $\endgroup$
    – Argyll
    Commented Sep 19, 2018 at 18:31
  • $\begingroup$ $F^{\mu\nu}$ is vector-like (all horizontal) as well as $F_{\mu\nu}$ (all vertical), so tracing over their outer product is equivalent to the inner product. $\endgroup$
    – user73236
    Commented Feb 2, 2021 at 17:11
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    $\begingroup$ I am surprised after all these years nobody has objected to the statement that $F_{\mu\nu}F^{\mu\nu}$ should be zero. It should not, and, since this was clearly originally the field tensor of electromagnetic field, $F_{\mu\nu}F^{\mu\nu}=0$ would amount to the action of electromagnetic field being equal 0 for any configuration of the field, since $\frac{-1}{4 \pi c}F_{\mu\nu}F^{\mu\nu}$ is the Lagrangian density of electromagnetic field. It is actually the difference between electric and magnetic energy density. $\endgroup$ Commented Dec 22, 2023 at 11:47

1 Answer 1

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I try to answer starting from the case of square matrices. There is some care to take while considering a "hidden" isomorphism of vector spaces. In any case, let $V$ be a finite dim. vector spaces over a field $\mathbb K$ (for simplicity $\mathbb R$ ), with basis $\{e_i\}$ of cardinality $n$.

It is well known that there exists an isomorphism of vector spaces $$\Phi:\operatorname{Hom}_\mathbb K(V,V)\rightarrow V^{*}\otimes V, $$

with $$\Phi(\phi)=a_{ij}f_i\otimes e_j,$$ where $\phi\in \operatorname{Hom}_\mathbb K(V,V)$ and $\phi(e_i):=a_{ij}e_j$ for all $i,j=1,\dots,n$. $\{f_i\}$ is the dual basis on $V^{*}$ of the basis $\{e_i\}$ on $V$, i.e. $f_i(e_j)=\delta_{ij}$.

We use the Einstein convention for repeated indices.

We know how to define the trace operator $\operatorname{Tr}$ on the space $\operatorname{Hom}_\mathbb K(V,V)$; the trace is computed on the square matrix representing each linear map in $\operatorname{Hom}_\mathbb K(V,V)$. Let us move to the r.h.s. of the isomorphism $\Phi$.

  • trace operator on $V^{*}\otimes V$

Let $$\operatorname{Tr}_1: V^{*}\otimes V\rightarrow \mathbb K, $$

be given by $\operatorname{Tr}_1(g\otimes v):=g(v)$.

Lemma $\operatorname{Tr}_1$ is linear and satisfies $$\operatorname{Tr}_1\circ \Phi=\operatorname{Tr}.$$

proof: just use definitions.

  • trace operator on $(V^{*}\otimes V)\otimes\dots\otimes (V^{*}\otimes V) $

Using the $n=1$ case we introduce

$$\operatorname{Tr}_n: \underbrace{(V^{*}\otimes V)\otimes\dots\otimes (V^{*}\otimes V)}_{n-\text{times}} \rightarrow \mathbb K, $$

with $\operatorname{Tr}_n(f_1\otimes v_1\otimes\dots\otimes f_n\otimes v_n):=\prod_{i=1}^n f_i(v_i)$.

Lemma $\operatorname{Tr}_n$ is linear and invariant under permutations on $(V^{*}\otimes V)^{\otimes n}$; it satisfies $$\operatorname{Tr}_n\left(\Phi(\phi_1)\otimes\dots\otimes\Phi(\phi_n)\right)=\prod_{i=1}^n \operatorname{Tr}(\phi_i), $$ for all $\phi_i\in \operatorname{Hom}_\mathbb K(V,V)$.

proof: we prove the second statement. We introduce the notation $$\Phi(\phi_k):= a^k_{i_kj_k}f_{i_k}\otimes e_{i_k}\in V^{*}\otimes V,$$ for all $k=1,\dots,n$. We arrive at $$\operatorname{Tr}_n\left( (a^1_{i_1j_1}f_{i_1}\otimes e_{i_1})\otimes\dots\otimes (a^n_{i_nj_n}f_{i_n}\otimes e_{i_n})\right)=a^1_{i_1j_1}\dots a^n_{i_nj_n}f_{i_1}(e_{i_1})\dots f_{i_n}(e_{i_n})=\text{remember the definition of dual basis}= a^1_{i_1j_1}\dots a^n_{i_nj_n}\delta_{i_1j_1}\dots\delta_{i_nj_n}= a^1_{i_1i_1}\dots a^n_{i_ni_n}\\=\prod_{i=1}^n \operatorname{Tr}(\phi_i),$$ as claimed.

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  • $\begingroup$ @Argyll did it help? $\endgroup$
    – Avitus
    Commented Jul 28, 2014 at 18:02
  • $\begingroup$ $a_{ij}$ should be changed to be $a_{ji}$ @Avitus $\endgroup$
    – yang
    Commented Sep 7, 2014 at 3:21
  • $\begingroup$ @Avitus: Yes it does. Thank you. But I'm still not sure about the connection between contraction and tensor product or any mathematical objects. $\endgroup$
    – Argyll
    Commented Sep 16, 2014 at 1:13
  • $\begingroup$ In which sense? Maybe you can extend your OP a bit $\endgroup$
    – Avitus
    Commented Sep 16, 2014 at 7:47
  • $\begingroup$ Good answer, though $$\Phi(\phi_k):= a^k_{i_kj_k}f_{i_k}\otimes e_{i_k}\in V^{*}\otimes V,$$ should be $$\Phi(\phi_k):= a^k_{i_kj_k}f_{i_k}\otimes e_{j_k}\in V^{*}\otimes V,$$ and thereon afterwards. $\endgroup$
    – TheDawg
    Commented Apr 26, 2022 at 13:11

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