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What I know: (1) If $V$ is a finite dimensional vector space then there is a natural isomorphism from $V$ to its double dual $V^{**}$. (2) There is no natural isomorphism from $V$ to its dual $V^*$. (3) If we consider finite vector spaces with an inner product, then there is a natural isomorphism from $V$ to $V^*$.

Intuitively, naturality can in all these 3 scenarios be understood as without need for some additional structure.

Naturality can also be made more precise in all 3 scenarios in a categorical sense:

  1. Define suitable categories (eg: vector spaces, vector spaces with inner products)
  2. Define functors (dualization, double dualization, identity functor)
  3. Check for a natural transformation between these functors.

The intuitive concept of naturality in linear algebra can be nicely translated into categorical language.

According to Wikipedia, Symmetric Tensor there exists a natural isomorphism between the vector space of symmetric tensors of order $d$ over an $n$-dimensional vector space and the dual of the vector space ${\mathbb K}[X_1, \ldots, X_n]$ of homogeneous polynomials of degree $d$ in $n$ variables over $\mathbb K$.

What I do not know: How would I formulate naturality in this situation in a categorical sense?

My specific problem: I am not aware between which functors I should look for a natural transformation. I am looking for a full, precise definition of functors making clear all the details of types, variances and more.

Nobody-Knows-I-am-a-Dog
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    The vector space $\mathbb K_d[X_1,\ldots,X_n]$ is fixed, while the symmetric tensors of order $d$ over a $\mathbb K$-vector space with dimension $n$ depend on that vector space. Intuitively, the isomorphism is not natural, and probably involves choosing a basis. – SolubleFish May 08 '22 at 21:23
  • @N You might find my old answer here to be helpful – Ben Grossmann May 08 '22 at 22:18
  • What do you mean by "there exists no natural isomorphism between $V$ and $V^$, but it's true for Euclidean spaces"? One functor is covariant and the other is contravariant. In the category of Euclidean spaces and linear isometries, these two functors are "anti-naturally isomorphic". If you say they are "naturally isomorphic", it implies you are considering the functoriality on $V^$ which sends $f$ to $(f^†)^*$ where $f^†$ is the adjoint of $f$... so it's a bit confusing when you don't say how your functors act on morphisms. – Dabouliplop May 09 '22 at 00:51
  • (Note : the category of Euclidean spaces and all linear maps is equivalent to the category of finite dimensional real vector spaces, even if you need non-canonical choices to go in the other direction.) – Dabouliplop May 09 '22 at 00:51
  • @SolubleFish Agreed. I was sloppy in writing up my question. The symmetric tensors are naturally isomorphic to the dual of the vector space of homogeneous polynomials (source: Wikipedia: Symmetric Tensors). I'll edit the question. This still leaves open my question as to how define naturality in this context, independently of whether it holds true or not. – Nobody-Knows-I-am-a-Dog May 09 '22 at 15:06
  • @Dabouliplop Yes. I did not make explicit all the details of the motivating examples which you filled in. In my question I do not even see how to obtain the necessary functors for a natural transformation, yet obtain their variances. – Nobody-Knows-I-am-a-Dog May 09 '22 at 15:21

1 Answers1

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I think there is a categorical description which captures some of the issues you are alluding to. It is based on my reading of Riehl,$\textit{Category Theory in Context }$ Chp2. Here is the setup:

Fix a field $\mathbb{K}$ and let $Vect_{\mathbb{K}}$ be the category of vector spaces over $\mathbb{K}$, with linear morphisms and $Comm_{\mathbb{K}}$ commutative, associative, unital algebras over $\mathbb{K}$ with k-algebra morphisms. For $V\in Vect_{\mathbb{K}}$:

Let $\mathcal{T}(V) = \bigoplus_{k=0}^{\infty} V^{\otimes k}$.

Let $\mathcal{S}(V)$ be the symmetric algebra, ($\mathcal{S}(V) = \mathcal{T}(V)/\mathcal{I}$, where $\mathcal{I}=\langle x\otimes y - y \otimes x\rangle$).

Let $U: Comm_{\mathbb{K}} \rightarrow Vect_{\mathbb{K}}$ be the functor which forgets the algebra structure.

Let $G_n: Comm_{\mathbb{K}} \rightarrow Sets$ be the functor defined by $G_n(A) = \{(a_1,\dots,a_n) : a_i \in A\} $, with obvious action on morphisms.

Finally, define functor $F_V :Comm_{\mathbb{K}} \rightarrow Sets$ by $F_V(A) = Hom_{Vect}(V, U(A))$.

Both $F_V$ and $G_n$ are covariant in $A$, are representable functors and have universal objects,namely, $\mathcal{S}(V)$ for $F_V$ and $\mathbb{K}[X_1,\dots,X_n]$ for $G_n$. I.e. we have natural bijections: \begin{align*} & Hom_{Comm}(\mathcal{S}(V),A) \simeq Hom_{Vect}(V,U(A)) = F_V(A) & (1) \\ & Hom_{Comm}(\mathbb{K}[X_1,\dots,X_n],A) \simeq G_n(A) & (2) \end{align*} The bijection in (1) is given by: $f\in Hom_{Vect}(V,U(A)), f:V\rightarrow A \leftrightarrow \tilde{f}:(v_1, \dots ,v_n)\rightarrow f(v_1)\dots f(v_n) \in A$. $\tilde{f}:V\times \dots \times V \rightarrow A$ is multilinear and induces an algebra morphism $\mathcal{S}(V)\rightarrow A$. For (2), the bijection is given by $(a_1,\dots ,a_n) \leftrightarrow f:\mathbb{K}[X_1,\dots ,X_n]\rightarrow A$, where $f(X_i)=a_i$.

For (1), the bijection is natural in both $V$ and $A$, i.e. $\mathcal{S}$ is left adjoint to $U$.

Here is the point:

Let $n=\text{Dim}(V)$. Then there is a bijection between $\textit{sets}$ $F_V(A) \leftrightarrow G_n(A)$, as follows: fix an ordered basis $(v_1,\dots,v_n)$ for $V$. Then every $(a_1,\dots,a_n)\in G_n(A)$ defines a vector space homomorphism $V\rightarrow A$ by $v_i \rightarrow a_i$ and vica versa. So by the Yoneda lemma, this isomorphism in the category of $Sets$ implies an isomorphism between $\mathcal{S}(V)$ and $\mathbb{K}[X_1,\dots,X_n]$ in the $Comm_{\mathbb{K}}$ category. Furthermore, for fixed $V$, considering $F_V$ and $G_n$ as functors $Comm_{\mathbb{K}} \rightarrow Sets$, the bijection is natural. A different choice of basis will conjuagte the isomorphism of $\mathcal{S}(V)$ and $\mathbb{K}[X_1,\dots,X_n]$ by an automorphism of $V$.

Like many categorical arguments (at least for me) this either elucidates or obfuscates the issues. I hope you find it helpful.

ahersh23
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