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edited Jan 5 2010 at 18:52

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 so $f$ has a right-sided inverse. Similarly, $\eta\epsilon - \phi=-g*f$ yields that $g*f$ is $f$. \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.

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edited Jan 5 2010 at 18:44

darij grinberg
12.6k●1●24●75

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.

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edited Jan 5 2010 at 18:34

darij grinberg
12.6k●1●24●75

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 using $\Delta\left(C_n\right)\subseteq\sum_{i=0}^n C_i\otimes C_{n-i}$ 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 so is $f$. Theorem 1 is proven.

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answered Jan 5 2010 at 18:26

darij grinberg
12.6k●1●24●75

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 using $\Delta\left(C_n\right)\subseteq\sum_{i=0}^n C_i\otimes C_{n-i}$ 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 so is $f$. Theorem 1 is proven.