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.


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.


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?)

  • 1
    $\begingroup$ Now here's a question: If one answerer provides a proof, and another answerer provides a reference, who should get their answer accepted? $\endgroup$ Jan 5, 2010 at 19:00
  • $\begingroup$ can you please give me any reference where I can find a proof of the fact that the functor gr is symmetric monoidal? $\endgroup$
    – Victor TC
    Apr 15, 2019 at 15:34
  • $\begingroup$ @Victor I don’t know a reference off the top of my head, but the statement is not hard. If you post it as a question, I will try to work out the details. $\endgroup$ Apr 15, 2019 at 18:47
  • $\begingroup$ thank you. Indeed, I posted a similar question which would be resolved by proving gr is symmetric monoidal, please see math.stackexchange.com/questions/3143378/… $\endgroup$
    – Victor TC
    Apr 15, 2019 at 22:25

2 Answers 2


There is a known theorem which, if I correctly understand your question, answers it (because the Hopf algebra antipode is defined as the $\ast$-inverse of $\mathrm{id}$):

Theorem 1. Let $A$ be an algebra and $\left(C,\left(C_n\right)_{n\geq 0}\right)$ a filtered coalgebra, i. e. a coalgebra $C$ and a sequence $\left(C_n\right)_{n\geq 0}$ such that:

$C_n$ is a vector subspace of $C$ for every $n\geq 0$;

$C=\bigcup_{n\geq 0}C_n$;

$\Delta\left(C_n\right)\subseteq\sum_{i=0}^n C_i\otimes C_{n-i}$ for every $n\geq 0$.

Let $f:C\to A$ be a linear map such that the restriction $f\mid_{C_0}:C_0\to A$ is $\ast$-invertible. Then, $f$ itself is $\ast$-invertible.

Proof of Theorem 1. Since $f\mid_{C_0}:C_0\to A$ is $\ast$-invertible, there exists a map from $C_0$ to $A$ which is the $\ast$-inverse of $f\mid_{C_0}$. Let $g$ be an arbitrary linear extension of this map to the whole $C$. So $g:C\to A$ is a linear map such that $g\mid_{C_0}:C_0\to A$ is a $\ast$-inverse of $f\mid_{C_0}:C_0\to A$. In other words, $\left(f*g\right)\mid_{C_0}=\eta\epsilon\mid_{C_0}$ (sorry, I call $\eta$ what you denote by $e$). In yet other words, $\phi\mid_{C_0}=0$, where $\phi:C\to A$ is the linear map defined by $\phi = \eta\epsilon - f*g$. An easy induction (only using $\Delta\left(C_n\right)\subseteq\sum_{i=0}^n C_i\otimes C_{n-i}$ and $\phi\mid_{C_0}=0$) shows that $\phi^i\left(C_n\right)=0$ for every integers $i$ and $n$ satisfying $i>n\geq 0$, where $\phi^i$ means the $i$-th power of $\phi$ with respect to the convolution $\ast$. Thus, the map $\sum_{i=0}^{\infty} \phi^i:C\to A$ is well-defined (in fact, the sum $\sum_{i=0}^{\infty} \phi^i:C\to A$ converges pointwise, as $C=\bigcup_{n\geq 0}C_n$). But $\left(\eta\epsilon - \phi\right)\ast\left(\sum_{i=0}^{\infty} \phi^i\right)=\eta\epsilon$ (by the geometric series formula, since $\eta\epsilon$ is the unity of the ring $\mathrm{Hom}\left(C,A\right)$ with multiplication $\ast$) and $\left(\sum_{i=0}^{\infty} \phi^i\right)\ast\left(\eta\epsilon - \phi\right)=\eta\epsilon$ (for the same reason). Hence, $\eta\epsilon - \phi$ is $\ast$-invertible. But $\eta\epsilon - \phi=-f*g$ (by the definition of $\phi$). Thus, $-f*g$ is $\ast$-invertible. Hence, so is $f*g$, and thus $f$ has a right-sided inverse. Similarly, $g*f$ is $\ast$-invertible, and thus $f$ has a left-sided inverse. Therefore, $f$ is invertible. Theorem 1 is proven.

EDIT: Okay, let me explain how to get your assertion from Theorem 1: Since $\mathrm{gr} V$ is a Hopf algebra, the identity $\mathrm{id}:\mathrm{gr} V\to\mathrm{gr} V$ has a $\ast$-inverse. Hence, its restriction $\mathrm{id}\mid_{V_0}:V_0\to\mathrm{gr} V$ to the component $V_0$ of $\mathrm{gr} V$ also has a $\ast$-inverse. This $\ast$-inverse must have its image in $V_0$ (because otherwise we can chain it with the projection $\mathrm{gr} V\to V_0$ and get another $\ast$-inverse of $\mathrm{id}\mid_{V_0}:V_0\to\mathrm{gr} V$, but the $\ast$-inverse is unique when it exists, so it must be the same one). Hence, the map $\mathrm{id}\mid_{V_0}:V_0\to V_0$ has an $\ast$-inverse. Thus, the map $\mathrm{id}\mid_{V_0}:V_0\to V$ has an $\ast$-inverse. Now, Theorem 1 yields that so does $\mathrm{id}:V\to V$, and we are done.

  • $\begingroup$ Just to clarify, $\ast$ in your proof is the multiplication on $\Hom(C,A)$ given by $f\ast g = m \circ (f\otimes g) \circ \Delta$, where $m$ is the multiplication and $\Hom(C,A)$ is the space of all linear maps? $\endgroup$ Jan 5, 2010 at 18:36
  • $\begingroup$ Anyway, so then the conclusion of your answer is to let A = C = the filtered bialgebra, whence the map id: C_0 \to A is *-invertible because it's invertible in associated-graded? $\endgroup$ Jan 5, 2010 at 18:39
  • $\begingroup$ Yes for both of your comments. $\endgroup$ Jan 5, 2010 at 18:42
  • $\begingroup$ Ah, right, so for my own sake I'll spell out the argument pedantically. The map id: V_0 \to V_0 is *-invertible, because this is (the restriction of) the map in grV, and the inverse is the restriction of the antipode s:grV \to grV to V_0. But extend this restriction s_0: V_0 \to V_0 to s_0: V_0 \to V by the natural embedding; then it is a *-inverse of the natural embedding V_0 \to V. Thus, the map id: V\to V restricts to a *-invertible map on V_0, and now run the above theorem. $\endgroup$ Jan 5, 2010 at 18:56

There is a theorem by Takeuchi [Takeuchi, Mitsuhiro. Free Hopf algebras generated by coalgebras. J. Math. Soc. Japan 23 (1971), 561--582. MR0292876] that states that if $A$ is an algebra and $C$ is a coalgebra, then a map $f\in\hom(C,A)$ is convolution invertible iff its restriction $f|_{C_0}\in\hom(C_0,A)$ to the coradical $C_0$ of $C$ is convolution invertible. In particular, if $A=C$ is a bialgebra, to check that it is a Hopf algebra you need only check that $\mathrm{id}_{H_0}:H_0\to H$ is convolution invertible. If your filtration has $H_0\subseteq F_0$, and $F_0$ is a Hopf algebra, then this is automatic. Another simple instance of this is the classical case where $H_0=K$.

  • 2
    $\begingroup$ Great! I wonder why I failed to learn this theorem... And this is why "connected" bialgebras are Hopf, a comment that's been made two or three times without proof over at the n-category cafe. Anyway, I hope never to check that an antipode exists ever again :) $\endgroup$ Jan 5, 2010 at 18:59
  • $\begingroup$ Indeed, for a connected Hopf algebras you can use the coradical filtration to check, and the condition is the automatic. $\endgroup$ Jan 5, 2010 at 19:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.