# When is an exponential functor a bialgebra?

My question is about the notion of exponential functors as they are frequently defined in the literature on (strict) polynomial functors, e.g., the paper "General Linear and Functor Cohomology" by Franjou, Friedlander, Scorichenko, and Suslin, or the paper "Bar complexes and extensions of classical exponential functors" by Touze. The essentials of the definition are as follows:

Let $k$ be a field, and let $A$ be a functor from the category of finite-dimensional (f.d.) k-vector spaces to the category of k-vector spaces. The tensor product of two such functors is defined in the obvious way.

Given such a functor $A$, suppose there exists a natural transformations $k \rightarrow A$ (viewing $k$ here as a constant functor) and $m: A \otimes A \rightarrow A$ such that for each f.d. vector space $V$, $A(V)$ is made into an associative $k$-algebra. Now say that $A$ is an exponential functor if for each pair of f.d. vector spaces $V$ and $W$, the composite map

$A(V) \otimes A(W) \stackrel{A(\iota_V) \otimes A(\iota_W)}{\rightarrow} A(V \oplus W) \otimes A(V \oplus W) \rightarrow A(V \oplus W)$,

induced by the inclusions $\iota_V: V \rightarrow V \oplus W$ and $\iota_W: W \rightarrow V \oplus W$ and the multiplication in $A(V \oplus W)$, is an isomorphism of $k$-vector spaces. Examples of exponential functors include the functor $S$ that takes a vector space $V$ to the symmetric algebra $S(V)$, the functor $\Lambda$ that takes $V$ to the exterior algebra $\Lambda(V)$, or the functor $\Gamma$ that takes $V$ to the divided power algebra $\Gamma(V)$.

Now, it seems to be well-known in the area that if $A$ is an exponential functor in the above sense, then each $A(V)$ is automatically not just an algebra, but a $k$-bialgebra, with the coproduct defined as the composite map

$\Delta_A(V): A(V) \rightarrow A(V \oplus V) \rightarrow A(V) \otimes A(V)$,

where the first arrow is the map induced by the diagonal map $V \rightarrow V \oplus V$, and the second arrow is the exponential isomorphism $A(V \oplus V) \cong A(V) \otimes A(V)$. Details are provided in the literature for why $\Delta_A(V)$ is always coassociative, but never (that I have found) for why $\Delta_A(V)$ satisfies the remaining condition for $A(V)$ to be a bialgebra, i.e., $\Delta_A(V)$ should be an algebra homomorphism.

Why is $\Delta_A(V)$, as defined above, always an algebra homomorphism? Or, is it always an algebra homomorphism? What conditions on $A$ suffice to make $\Delta_A(V)$ always be an algebra homomorphism?

My question is a bit vague, because I have not told you how to view the tensor product $A(V) \otimes A(V)$ as an algebra; for the examples $A = S$ and $A = \Gamma$, it is the ordinary tensor product of algebras, while for $A = \Lambda$ it is the graded tensor product of algebras (the grading coming from the natural grading on $\Lambda$). So in general, the answer should probably depend on some kind of grading inherent to $A$.

One easy case where I can see that $\Delta_A(V)$ should be an algebra homomorphism is when each $A(V)$ is a (graded) commutative algebra, and $A(V) \otimes A(V)$ is viewed as the orindary (graded) tensor product of algebras, for then $\Delta_A(V)$ can be seen to be the composite of two algebra homomorphisms.

• As far as I can tell, $A$ necessarily factors through the forgetful functor from unital associative $k$-algebras to $k$-vector spaces. That is, a map of vector spaces induces a map of unital associative $k$-algebras. This seems to reduce your question to the question of whether the exponential isomorphism is an algebra map. – S. Carnahan Jul 30 '14 at 14:14
• Then the problem seems to reduce to knowing that the images of the maps $A(V) \rightarrow A(V \oplus W)$ and $A(W) \rightarrow A(V \oplus W)$ commute in some appropriate sense inside the algebra $A(V \oplus W)$. It's not clear to me at the moment whether or not this is automatic. – Christopher Drupieski Jul 30 '14 at 14:22
• Did you read section 5 of arXiv:0902.4459 ? It may be related. – Wilberd van der Kallen Jul 30 '14 at 14:37
• I've read Section 5 of arXiv:0902.4459 quite a bit, but perhaps I had been focusing on the wrong bits. I think Lemmas 5.4 and 5.10 combine to state that the condition I described in my previous comment is both necessary and sufficient. – Christopher Drupieski Jul 30 '14 at 14:47

Tensor porduct of functors between categories $\mathcal{V} \rightarrow \mathcal{W}$ can be defined as soon as $\mathcal{W}$ is a monoidal category. The functor category $[\mathcal{V}, \mathcal{W}]$ becomes a monoidal category under this tensor. A monoid $A$ in this monoidal category is the same as a functor $A : \mathcal{V} \rightarrow \mathrm{Mon}(\mathcal{W})$, where $\mathrm{Mon}(\mathcal{W})$ is the category of monoids in $\mathcal{W}$.
If $\mathcal{W}$ is symmetric (or just braided) monoidal category then $\mathrm{Mon}(\mathcal{W})$ becomes a monoidal category itself. In this case a bimonoid in $\mathcal{W}$ is defined to be a comonoid object in $\mathrm{Mon}(\mathcal{W})$. If $\mathcal{V}$ itself is monoidal and $A : \mathcal{V} \rightarrow \mathrm{Mon}(\mathcal{W})$ is an opmonoidal functor (this means that there are certain structural maps $A(X\otimes'Y) \rightarrow A(X)\otimes A(Y)$), then $A$ will take comonoids in $\mathcal{V}$ to comonoids in $\mathrm{Mon}(\mathcal{W})$, i.e. to bimonoids.
Suppose that $\mathcal{V}$ is monoidal under the direct sum. Then the functor $A : \mathcal{V} \rightarrow \mathrm{Mon}(\mathcal{W})$ becomes monoidal (a notion dual to opmonoidal) with the structural maps as defined in the question. If the monoidality structure morphisms are invertible, then $A$ also is opmonoidal with the inverse maps as the structural morphisms.
Every object of $\mathcal{V}$ has a canonical comonoid structure via the diagonal. Under the conditions explained $A : \mathcal{V} \rightarrow \mathrm{Mon}(\mathcal{W})$ will take these canonical comonoids to bimonoids in $\mathcal{W}$.
Edit: This to work, the monoidal structure maps (the exponential maps) have to be actually monoid morphisms. This holds when $A(X) \rightarrow A(X\oplus Y)$ and $A(Y) \rightarrow A(X\oplus Y)$ commute with each other in the monoid structure of $A(X\oplus Y)$ in the appropriate sense. That is what is pointed out in the comments to OP.