In this post, we will talk about canonical isomorphisms with non-canonical-looking inverses related to tensor products in linear algebra. I wrote this simple post to review some linear algebra and test MathJax.

Given finite dimensional vector spaces $U$ and $V$ over the same field $k$, let $\text{Hom}_k \,(U,V)$ denote the set of $k$-linear map (i.e. linear transformation) from $U$ to $V$. It is easy to see that $\text{Hom}_k \,(U,V)$ is itself a $k$-vector space. One might learn fun fact that we can identify $\text{Hom}_k \,(U,V)$ with the tensor product $U^\ast \otimes V$, where $U^\ast$ is the dual space of $U$ (i.e., $U^\ast = \text{Hom}_k \,(U,k)$). Precisely, there is a natural isomorphism $\phi : U^\ast \otimes V \rightarrow \text{Hom}_k \,(U,V)$ given by

\[f \otimes v \longmapsto (u \mapsto f(u)v),\]

for $v \in V$, $u \in U$, and $f \in U^\ast$. It is obvious that $\phi$ is a $k$-linear map. To show it is actually an isomorphism of $k$-vector spaces, we find its two-sided inverse. Let $\lbrace u_i \rbrace_{i \in I}$ be a basis for $U$ and $\lbrace u_i^\ast\rbrace_{i\in I}$ be the corresponding dual basis for $U^\ast$; that is, $u_i^\ast(u_j) \coloneqq \delta_{ij}$ where $\delta_{ij}$ is Kronecker’s delta.

Then we define a $k$-linear map $\psi : \text{Hom}_k \,(U,V) \rightarrow U^\ast \otimes V$ as

\[g \longmapsto \sum_{i \in I}u_i^\ast \otimes g(u_i),\]

for $g \in \text{Hom}_k \,(U,V)$. It is easy to check $\psi$ is actually a $k$-linear map and the two-sided inverse of $\phi$. One may feel awkward since our construction of $\psi$ seems to highly depend on the choice of $\left\{ u_i \right\}_{i \in I}$. However, surprisingly, $\psi$ is also canonical since its inverse does not depend on the choice of basis. This shows it is possible that a $k$-linear map essentially does not depend on the choice of basis, though its construction or expression involves specific elements of some basis. Otherwise, one may show expliclity that for another basis $\{t_i\}_{i \in I}$ for $U$ and the correponding dual basis $\{t_i^\ast\}_{i \in I}$, new $k$-linear map $\psi’ : \text{Hom}_k \,(U,V) \rightarrow U^\ast \otimes V$ defined using $\{t_i^\ast\}_{i \in I}$ coincide with $\psi$. It is quite hard (impossible for me) to imagine constructing a two-sided inverse of $\phi$ without dual bases. However, it does not imply $\phi$ is non-canonical.

Let’s move on to another example. Let $V$ be a vector space over $k$, and $V^{\otimes n} \coloneqq \underbrace{V \otimes \cdots \otimes V}_{n\text{-times}}$ and $V^{\otimes 0} \coloneqq k$. If $I = (i_1, \ldots, i_n)$ and $J = (j_1, \ldots, j_n)$ are two sequences of $n$ distinct indices from $\{1, \ldots, p\}$ and $\{1, \ldots, q\}$, respectively, we define a contraction map

\[c^I_J : V^{\otimes p} \otimes (V^\ast)^{\otimes q} \longrightarrow V^{\otimes (p-n)} \otimes (V^\ast)^{\otimes (q-n)}\]

which sends $v_1 \otimes \cdots \otimes v_p \otimes f_1 \otimes \cdots \otimes f_q$ to

\[\left( \prod_{\alpha=1}^n f_{j_\alpha}(v_{i_\alpha}) \right) v_1 \otimes \cdots \otimes \overbracket{v_{i_1}\otimes \cdots \otimes v_{i_n}} \otimes \cdots \otimes v_p \otimes f_1 \otimes \cdots \otimes \overbracket{f_{j_1}\otimes \cdots \otimes f_{j_n}} \otimes \cdots \otimes f_q\]

where $\overbracket{\hspace{10pt}}$ means omitting the terms below. If $p=q=n$ and $I=J=(1, \ldots, n)$, then the contraction becomes a map $c : V^{\otimes n} \otimes (V^\ast)^{\otimes n} \to k$ given by

\[v_1 \otimes \cdots \otimes v_n \otimes f_1 \otimes \cdots \otimes f_n \longmapsto \prod_{\alpha =1}^n f_\alpha(v_\alpha).\]

Then $c$ induces a $k$-linear map $c_\star : (V^\ast)^{\otimes n} \to (V^{\otimes n})^\ast$ defined as

\[f_1 \otimes \cdots \otimes f_n \longmapsto (v_1 \otimes \cdots \otimes v_n \mapsto c(v_1 \otimes \cdots \otimes v_n \otimes f_1 \otimes \cdots \otimes f_n)).\]

The construction of $c$ and $c_\star$ are natural. We claim that $c_\star$ is actually an isomorphism; that is, we can identify $(V^\ast)^{\otimes n}$ with $(V^{\otimes n})^\ast$ (which sounds intuitive!). Let $\{e_1, \ldots, e_m\}$ be a basis for $V$, and $\{e_1^\ast, \ldots, e_m^\ast\}$ be the corresponding dual basis for $V^\ast$. Then $\{e_{i_1} \otimes \cdots \otimes e_{i_n} \}_{\underline{i} \in I^n}$ forms a basis for $V^{\otimes n}$, where $I = \{1, \ldots, m\}$ and $\underline{i} \coloneqq (i_1, \ldots, i_n) \in I^n$. We also have the correponding dual basis $\{\left(e_{i_1} \otimes \cdots \otimes e_{i_n}\right)^\ast \}_{\underline{i} \in I^n}$ for $(V^{\otimes n})^\ast$. Define a $k$-linear map $d_\star : (V^{\otimes n})^\ast \to (V^\ast)^{\otimes n}$ by

\[F = \sum_{\underline{i} \in I^n}\lambda_{\underline{i}}\left(e_{i_1} \otimes \cdots \otimes e_{i_n}\right)^\ast \longmapsto \sum_{\underline{i} \in I^n}\lambda_{\underline{i}}e_{i_1}^\ast \otimes \cdots \otimes e_{i_n}^\ast.\]

By tedious computations, one can show $d_\star$ is actually the two-sided inverse of $c_\star$. Similar to the previous example, the construction of $d_\star$ looks dependent on the choice of ${e_1, \ldots, e_n}$, but it does not in fact. Therefore, we can naturally identify $(V^{\otimes n})^\ast$ with $(V^\ast)^{\otimes n}$, though the expression of $d_\star$ involves ${e_1, \ldots, e_n}$. (Can you show that this identification may fail if $V$ is not finite-dimensional?)

In linear algebra, there are lots of objects with basis-free definition looking dependent to the choice of basis. Important examples taught in typical undergraduate linear algebra courses are determinant and trace. We define the determinant and trace of linear transformation $T$ using their matrix representation (i.e., using bases), but the values do not be affected by the choice of basis. Under the same philosophy, it is important to notice that a non-canonical-looking expression does not imply it is non-canonical.


References

  1. C. Curtis, Linear Algebra: An Introductory Approach, Springer-Verlag, 1984
  2. W. Fulton and J. Harris, Representation theory: a first course, Springer-Verlag, 2004