How do we define an abstract basis-independent trace operation?

Question

In this wikpedia page on Abstract index notation for tensors, it states that $t_{ab}{}{}^b$ denotes an "abstract basis-independent trace operation". I understand that this index notation doesn't refer to numerical indices and is basis independent. However, I haven't been able to find a definition of the abstract basis-independent trace operation. E.g. it seems that this is something unrelated.

Answer 1

Given any vector space $V$ over a field $\Bbb{F}$, we can always consider the natural evaluation mapping $\epsilon:V\times V^*\to\Bbb{F}$, given as $\epsilon(v,f):=f(v)$. Then, $\epsilon$ is clearly bilinear, so by the universal property of tensor products (which is in fact one of the cleanest, but abstract, ways to define the tensor product), there is a unique linear mapping $\epsilon':V\otimes V^*\to\Bbb{F}$ such that for all pure tensors $v\otimes f$, we have $\epsilon'(v\otimes f)=\epsilon(v,f)=f(v)$ (i.e such that 'the obvious' diagram commutes). This is the definition of the contraction/trace mapping.

By reasoning similarly, given vector spaces $V,V_1,\dots, V_k$ over a field $\Bbb{F}$, we have the linear mapping $\text{tr}:V\otimes V^*\otimes (V_1\otimes \cdots \otimes V_k)\to V_1\otimes \cdots \otimes V_k$ such that for all pure tensors, \begin{align} \text{tr}(v\otimes f\otimes v_1\otimes\cdots \otimes v_k)&=f(v)\cdot\,v_1\otimes\cdots \otimes v_k \end{align} Typically, the $V_i$'s are one of $V$ or $V^*$. So, whenever you have a tensor product space whereby in the factors, a vector space and its dual both appear, you can define a trace/contraction over those slots.

Now, we can ask the question of how to define the trace of an endomorphism $T:V\to V$. Well, assuming $V$ is finite-dimensional, one can prove (using a basis) that the natural mapping $V\otimes V^*\to \text{End}(V)$ (whose action on pure tensors is $v\otimes f$ being mapped to the endomorphism $x\mapsto f(x)\cdot v$) is actually an isomorphism. In this case, the trace/contraction mapping on $V\otimes V^*$ induces a corresponding map on $\text{End}(V)$. This is precisely the trace of an endomorphism as defined in linear algebra (i.e take a matrix representation relative to some fixed basis, and then sum the diagonal entries).