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Denote by $Vect^{fin}_{\mathbb{C}}$ the category of finite dimensional complex vector spaces. We will call a pair $(F,\tau)$ that consists of a functor $F \colon Vect^{fin}_{\mathbb{C}} \to Vect^{fin}_{\mathbb{C}}$ and a natural isomorphism $$ \tau_{V,W} \colon F(V \oplus W) \to F(V) \otimes F(W) $$ an exponential functor if $\tau$ satisfies the following associativity condition: $$ (\tau_{V,W} \otimes id_{F(X)}) \circ \tau_{V \oplus W, X} = (id_{F(V)} \otimes \tau_{W,X}) \circ\tau_{V,W \oplus X} $$ An example of such a functor is the full exterior power $\Lambda^*$ together with the natural iso $\Lambda^*(V \oplus W) \cong \Lambda^*(V) \otimes \Lambda^*(W)$. Another example along the same lines is $$ F_X(V) = \Lambda^*(V \otimes X) $$ for a fixed finite dimensional complex vector space $X$. Since we insist on finite dimensional vector spaces as our target category for $F$, the full symmetric power is not an example.

Are there any other examples of exponential functors on $Vect^{fin}_{\mathbb{C}}$? Is there a classification of exponential functors?

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    $\begingroup$ There almost certainly is a classification. I may try elaborate on this another time, but three things spring to mind: polynomials functors; lambda-rings; natural isomorphism $\Bbb{C}^n\simeq \oplus_{i=1}^n \Bbb{C}$. $\endgroup$ Apr 29, 2017 at 4:21
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    $\begingroup$ Do you want $F(0)=\mathbb{C}$ (i.e., $F(\bigoplus_iV_i)=\bigotimes_iF(V_i)$ for all finite direct sums, even the empty one)? Otherwise there's always the zero functor ... $\endgroup$ Apr 29, 2017 at 7:57
  • $\begingroup$ I asked a similar question here math.stackexchange.com/questions/1618100/… $\endgroup$ Apr 29, 2017 at 20:36
  • $\begingroup$ @JeremyRickard Indeed, I want $F(0) = \mathbb{C}$. $\endgroup$ Apr 29, 2017 at 23:01
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    $\begingroup$ But then you probably also want a coherence condition for $\tau_{0,W}$ and $\tau_{V,0}$, and also a symmetry condition for the $\tau_{V,W}$. I assume that you want $F$ to be a symmetric monoidal functor $(Vect^{fin}_{\mathbb{C}},\oplus,0) \to (Vect^{fin}_{\mathbb{C}},\otimes,\mathbb{C})$. I have done some google research. The term of an exponential functor is quite well-known in the world of polynomial functors. The definition there is a little bit different, but not much. It astounds me that I haven't been able to find a non-classical example mentioned anywhere. $\endgroup$
    – HeinrichD
    May 14, 2017 at 18:29

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In case anyone is still interested in this question:

There is a classification of polynomial exponential functors on the category $\mathcal{V}$ of finite-dimensional inner product spaces in terms of involutive $R$-matrices (i.e. involutive solutions to the Yang-Baxter equation). Basics about polynomial functors can be found in the book "Symmetric Functions and Hall Polynomials" by Macdonald (Chapter 1, Appendix A).

A functor $F \colon \mathcal{V} \to \mathcal{V}$ is polynomial, if for all linear maps $f_i \colon V \to W$ the result of $F(\lambda_1 f_1+ \dots + \lambda_n f_n)$ is a polynomial in the $\lambda_i$ with coefficients in $\hom(V,W)$. Each such functor has a direct sum decomposition into homogeneous components $F_n \colon \mathcal{V} \to \mathcal{V}$. Each homogeneous component has a linearisation $L_{F_n} \colon \mathcal{V}^n \to \mathcal{V}$ and $L_{F_n}(V_1, \dots, V_n)$ is defined to be the direct summand of $F_n(V_1 \oplus \dots \oplus V_n)$, on which $\lambda_1 id_{V_1} \oplus \dots \oplus \lambda_n id_{V_n}$ acts via multiplication by $\lambda_1 \cdot \lambda_2 \cdots \lambda_n$.

For an exponential functor $F$, the linearisation $L_{F_n}$ of its $n$-homogeneous summand $F_n$ is equivalent to $F_1^{\otimes n}$. Let $W = F_1(\mathbb{C})$ and $\tau \colon \mathbb{C}^2 \to \mathbb{C}^2$ be the isomorphism permuting the summands. Then $F_2(\tau) \colon F_2(\mathbb{C}^2) \to F_2(\mathbb{C}^2)$ restricts to a linear transformation $R \colon W^{\otimes 2} \to W^{\otimes 2}$ of the linearisation of $F_2$ that is involutive and satisfies the Yang-Baxter equation on $W^{\otimes 3}$. This can be extended to the statement that polynomial exponential functors up to natural equivalence of monoidal functors are in $1:1$-correspondence with certain equivalence classes of involutive $R$-matrices.

This is written up here. This result is used there to construct twists of complex $K$-theory localised at an integer over $SU(n)$, which was my original motivation for this question.

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Are there any other examples of exponential functors on $Vect^{fin}_{\mathbb{C}}$?

Here is a well-known explicit example that is closely related to $\Lambda^*$, but is not a special case of $F_X$: $$ F(V)=\det(V) $$ (i.e. the highest exterior power of $V$), with the natural isomorphism $$ \det(V\oplus W)\cong \det(V)\otimes\det(W). $$

Also, if $F$ is an exponential functor, then so is $F^{\otimes k}$ given by $$ F^{\otimes k}(V)=F(V)^{\otimes k} $$ and the corresponding $\tau^{\otimes k}$, for any $k\geq 0$. In particular, $\det^{\otimes k}$ and $F_X^{\otimes k}$ form larger families of exponential functors that deform $\det$ and $F_X$ (corresponding to $k=1$).

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    $\begingroup$ I don't see how to make a functor from the highest exterior power. What happens to maps between vector spaces of different dimensions? $\endgroup$ Apr 29, 2017 at 7:51
  • $\begingroup$ Jeremy, $\det $ of a map between vector spaces of different dimensions is $0$. $\endgroup$ Apr 29, 2017 at 13:05
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    $\begingroup$ You are right. I was about to say that existence of retractions is a problem. $\endgroup$ Apr 29, 2017 at 13:18
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    $\begingroup$ However, det seems to be a functor if we restrict the category to the subgroupoid, which contains just the isomorphisms. For the application I have in mind, this is probably enough. $\endgroup$ Apr 29, 2017 at 21:20
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    $\begingroup$ @UlrichPennig, you should clarify if you are interested in the groupoid in the question! The aswers are probably quite different. $\endgroup$ Aug 13, 2017 at 19:06

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