Coordinate-Free Definition of Trace, revisited.

I have some questions about a coordinate free definition of the trace of linear operators. This questions has been asked before in this forum (see [1,2]), but I haven't found the answers of my interest to be clear enough, so I will ask some further questions about the answer which I will enumerate them as Q.i), Q.ii), ... while I ask them.

Let me begin fixing the notation. Let $$V$$ be a vector space, $$V^*$$ its dual, $$V\otimes V^*$$ their tensor product and $$End(V)$$ the vector space of linear endomorphisms in $$V$$ (i.e. linear operators from $$V$$ to $$V$$).

As far as I understood, the answer begins stating that the mapping $$v\otimes w^*\to (u \to w^*(u)v)$$ for all $$v,u\in V$$ and $$w^*\in V^*$$ is a linear isomorphism. My first question is about this mapping. Q.i) What happens to those elements of $$V\otimes V^*$$ that are not pure, i.e. that are of the form $$\sum_i v_i\otimes w_i^*$$?

Later in the answer, a scalar $$w^*(v)$$ is associated to each element of the form $$v\otimes w^* \in V\otimes V^*$$. As far as I understood, is this scalar what is identified with the trace. Q.ii) What happens when we consider non pure tensor products like $$\sum_i v_i\otimes w_i^*$$? Q.iii) In this case, does the scalar takes the form $$\sum_i w^i(v_i)$$? Q.iv) Given, that the representation $$\sum_i v_i\otimes w_i^*$$ is not unique, why $$\sum_i w^i(v_i)$$ remains invariant?

EDIT 1: I changed not factorizable by not pure as suggested by @TrevorGunn.

• If $\phi$ is any linear map defined on a tensor product $V \otimes W$, then $\phi \big( \sum_i v_i \otimes w_i \big) = \sum_i \phi(v_i \otimes w_i)$. Commented Apr 3, 2021 at 23:25
• They're called "pure" tensors not "factorizable." Commented Apr 3, 2021 at 23:48
• So, let $\phi:V\otimes V^*\to End(V)$ be such that, for any $v\otimes w^*\in V\otimes V^*$, then $\phi(v\otimes w^*)u = w^*(u)v$ for all $u\in V$. How do you show this mapping is linear from the definition? More specifically, how do you evaluate $\phi(v\otimes w^*+k\, x\otimes y^*)$ for any $v\otimes w^*,x\otimes y^*\in V\otimes V^*$ and $k\in\mathbb{K}$ from the definition? Commented Apr 3, 2021 at 23:54
• I think most of your problems would be solved by the universal property of tensor products: for every bilinear map $\beta :U\times V\to W$, there exists a unique linear map $T:U\otimes V\to W$, such that $T(u\otimes v)=\beta (u,v)$. This result is the best way to construct a linear map defined on a tensor product.
– Ruy
Commented Apr 4, 2021 at 0:50
• The trace of an element in $V\otimes V^*$ can be defined in a coordinate-free way as follows: First, $End(V^*)$ contains a distinguished element known as the identity map $I$. Since there is a natural isomorphism $End(V^*) \simeq V^*\otimes V$, there is the corresponding distinguished element $\delta \in V^*\otimes V$. Since $V^*\otimes V$ is naturally isomorphic to $(V\otimes V^*)^*$, you can define the trace of $\phi \in V\otimes V^*$ to be $\operatorname{tr} \phi = \langle \delta, \phi\rangle$. Commented Apr 4, 2021 at 3:41

Q.i) The map is first specified by its action on separable elements. Since every element in general can be written as a linear combination of separable elements, the action of the map on a general element is given by linear extension. Explicitly, elements of the form $$\sum_{i}v_i \otimes w_i^*$$ are taken to maps of the form $$u \mapsto\sum_i w_i^*(u)v_i.$$

Q.ii and Q.iii) Yes, this scalar is the trace. Specifically, it is the trace of the element $$v\otimes w^*$$, or if you'd prefer, the linear map on $$V$$ that this element represents. The trace of a general (non-separable) element again has to be given through linearity. Using the same example as before, an element of the form $$\sum_{i}v_i \otimes w_i^*$$ has a trace of $$\mathrm{Tr}\left(\sum_i v_i \otimes w_i^*\right)=\sum_i \mathrm{Tr}\left(v_i \otimes w_i^*\right) = \sum_i w_i^*(v_i),$$ which is exactly what you suggested.

Q.iv) This comes down to whether the map is well-defined or not. This has to do with whether or not there actually exists a linear map $$\mathrm{Tr}$$ such that $$\mathrm{Tr}(v\otimes w^*) = w^*(v)$$ for all separable elements.

In general, you have to check the consistency of the map anytime its defined on a linearly dependent set (i.e., the set of separable tensors). The easiest way to show consistency is to explicitly construct such a map, using a basis for example. Fix a basis $$e_i \otimes e_j^*$$ for $$V\otimes V^*$$ and define the trace by $$\mathrm{Tr}(e_i\otimes e_j) = e^*_j(e_i) = \delta_{ij},$$ where $$\delta_{ij}$$ denotes the Kroencker delta. Note that there is no ambiguity in this definition here because we are first defining it on basis elements, and every general element has a unique decomposition in terms of basis vectors.

Given $$v = \sum_{i}v_ie_i$$ and $$w^* = \sum_{j}w_j^*e^*_j$$ (I am being a bit sloppy in distinguishing vectors and their components, but hopefully it's clear from context), we have: $$\mathrm{Tr}(v\otimes w^*) = \sum_{i,j}v_iw_j^*\mathrm{Tr}(e_i\otimes e_j^*) = \sum_{i}v_iw_i^* = w^*(v)$$

This shows that there exists a well-defined linear map such that $$\mathrm{Tr}(v \otimes w^*) = w^*(v)$$ for all separable elements.

• So the map $Tr$ is linear by assumption, and then it is fully determined by how it is defined to operate on the factorizable elements of $V\otimes V^*$? What puzzled me was that, in the end, to proof stuff, you need to resort to some basis. Commented Apr 4, 2021 at 0:04
• @JuanI.Perotti Yes, that's right. Definitions are conceptual, and that's what's important. The fact that you need to resort to something more concrete like a basis to actually prove things is quite typical. For example, it's common to define the tensor product $V\otimes W$ itself by a universal property, but you still need to show existence in the end, and that's done through a very concrete free product construction.
– EuYu
Commented Apr 4, 2021 at 3:30