Warning: older texts use the word "Hopf algebra" for what's now commonly called "bialgebra", whereas now "Hopf" is an extra condition. So as to avoid any confusion, I'll give my definitions before concluding with my question.

## Definitions

Let $C$ be a category with symmetric monoidal structure $\otimes$ and unit $1$ (and either strictify, or decorate all the following equations with associators and unitators and so on). An (associative, unital) *algebra* in $(C,\otimes)$ is an object $V$ along with maps $e: 1\to V$ and $m: V\otimes V \to V$ satisfying associativity and unit axioms: $m\circ(m\otimes \text{id}) = m\circ (\text{id}\otimes m)$ and $m\circ (\text{id}\otimes e) = \text{id} = m\circ (e\otimes \text{id})$. A (coassociative, counital) *coalgebra* is an object $V$ along with maps $\epsilon: V\to 1$ and $\Delta: V \to V\otimes V$ satisfying coassociativity and counit axioms. A *bialgebra* is any of the following equivalent things:

- A coalgebra in the category of algebras and algebra-homomorphisms ($1$ has its canonical algebra structure coming from the $\otimes$ axioms that $1\otimes 1 = 1$; in the tensor product of algebras, elements in the different multiplicands commute)
- An algebra in the category of coalgebras and coalgebra-homomorphisms
- An object $V$ with maps $e,m,\epsilon,\Delta$ satisfying the axioms above and a compatibility axiom: $$ \Delta \circ m = (m\otimes m) \circ (\text{id} \otimes \text{flip} \otimes \text{id}) \circ (\Delta \otimes \Delta) $$

A bialgebra can have the property of being *Hopf* (it is a property, not extra data): a bialgebra $V$ is *Hopf* if there exists an *antipode* map $s: V\to V$ satisfying
$$ m \circ (s\otimes \text{id}) \circ \Delta = e\circ \epsilon = m \circ (\text{id} \otimes s) \circ \Delta $$
Naturally, it's better to see these definitions than read them; check e.g. the Wikipedia article. If an antipode exists for a bialgebra, it is unique (justifying considering Hopfness a property rather than a structure) and it is an antihomomorphism for both the algebra and coalgebra structures.

Let VECT be the category of vector spaces (over your favorite field), with $\otimes$ the usual tensor product and $1$ the ground field. A ($\mathbb N$-)*filtered vector space* is a sequence $V = \{V_0 \hookrightarrow V_1 \hookrightarrow V_2 \hookrightarrow \dots\}$ in VECT. A morphism of filtered vector spaces $V \to W$ is a sequence of morphisms $V_n \to W_n$ so that every square commutes: $\{V_n \hookrightarrow V_{n+1} \to W_{n+1}\} = \{V_n \to W_n \hookrightarrow W_{n+1}\}$. Equivalently, a *filtered vector space* is a space $V \in $VECT along with an increasing sequence of subspaces $V_0 \subseteq V_1 \subseteq \dots \subseteq V$ such that $V = \bigcup V_n$, and a linear map of filtered vector spaces $V \to W$ is *filtered* if the image of $V_n$ lies in $W_n$ for each $n$.

Because $\otimes$ is exact in VECT (because every monomorphism splits), to a pair $V,W$ of filtered vector spaces we can define an $\mathbb N^2$-filtered space with $(p,q)$-part $V_p\otimes W_q$, and then we can define the $\mathbb N$-filtered space $V\otimes W$ by setting $(V\otimes W)_n$ to be the colimit of the diagram given by all $V_p\otimes W_q$ with $p+q \leq n$. Equivalently, we can take the tensor product in VECT of the unions $V = \bigcup V_n$ and $W = \bigcup W_n$, and then filter it by declaring that the $n$th part is the union of the $(p\otimes q)$th parts for $p+q = n$.

A ($\mathbb N$-)*graded* vector space is a sequence $\{V_0,V_1,V_2,\dots\}$ in VECT, or equivalently a space $V$ along with a direct sum decomposition $V = \bigoplus V_n$. A morphism of graded vector spaces preserves the grading.

Let $V$ be a filtered vector space. Its *associated graded* space $\text{gr}V$ is given by $(\text{gr}V)_n = V_n / V_{n-1}$, where $V_{-1} = 0$, of course. Then $\text{gr}$ is a symmetric monoidal functor, and so takes filtered bialgebras to graded bialgebras.

## Question

Let $V$ be a filtered bialgebra, i.e. a bialgebra in the category of filtered vector spaces. Then $\text{gr}V$ is a graded bialgebra. Suppose that $\text{gr}V$ is Hopf. Does it follow that $V$ is Hopf? I.e. suppose that $\text{gr}V$ has an antipode map. Must $V$ have an antipode map?

(Or perhaps it requires additional hypotheses, e.g. that we be in characteristic 0, or that $V$ is *locally finite* in the sense that each $V_n$ is finite-dimensional?)

gris symmetric monoidal? $\endgroup$ – Victor TC Apr 15 '19 at 15:34gris symmetric monoidal, please see math.stackexchange.com/questions/3143378/… $\endgroup$ – Victor TC Apr 15 '19 at 22:25