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Hi Darij,

A geometric way of thinking about the shuffle algebra

A geometric way of seeing $T(V)$ with the shuffle product is by considering functions on the loop space of $V^*$ (i.e. the space of continuous maps from $S^1$ to $V^*$) that are given by iterated integrals. You'll see that the product of two iterated integrals is precisely the shuffle product. Moreover, this way you can see the coproduct coming from the concatenation of loops and the antipode from reversing orientation (if I remember well).

But here you have to be in a situation when $V^{**}\cong V$.

Let me be a bit more precise. For convenience I will work with $T(V^*)$ equipped with the shuffle product $\star$, the deconcatenation coproduct $\Delta$, and $S$ as you defined it.

Now let me consider the algebra $\mathcal A$ of functionals on $L(V^*)=C_*^0(S^1,V^*)$ (the subscript $*$ means that I ask that $0$ is sent to $0$). There is an algebra monomorphism $T(V^*)\to \mathcal{A}$ given as follows: $$ \xi_1\otimes\cdots\otimes \xi_n\mapsto \left(\gamma\mapsto \int_{0<t_1<\cdots<t_n<1}\xi_1(\gamma(t_1))\dots\xi_n(\gamma(t_n))dt_1\dots dt_n\right) $$

Now observe that the composition of loops and the orientation reversing define algebra morphisms $\Delta_A:\mathcal A\to\mathcal A\otimes\mathcal A$ and $S_A:\mathcal A\to\mathcal A$.

The point is that $\Delta_A$ and $S_A$ do not really satisfy the axioms you want (e.g. coassociativity of $\Delta_A$) BUT their restriction onto the image of $T(V^*)$ does (you have to use an avatar of Stokes' Theorem to see this - or, shortly: iterated integrals are not sensitive to reparametrization), and actually coincide with $\Delta$ and $S$ (very simple computation).

The main point concerning the antipode is that

any connected filtered bialgebra is a filtered Hopf algebra, the antipode being defined as $S(x)=\sum_{k\geq0}(\eta\circ\epsilon-id)^{*k}(x)$

Here $*$ denotes the convolution product. In your example the bialgebra you consider is actually graded so the result applies. You can find the above claim (and its proof) in these lecture notes (I think you are going to like them) by Dominique Manchon (Corollary II.3.2).

The group algebra of a free group

The degree completion of $T(V)$ is the structure ring of the pro-unipotent completion of a free group.

I hope this can help.

Hi Darij,

A geometric way of thinking about the shuffle algebra

A geometric way of seeing $T(V)$ with the shuffle product is by considering functions on the loop space of $V^*$ (i.e. the space of continuous maps from $S^1$ to $V^*$) that are given by iterated integrals. You'll see that the product of two iterated integrals is precisely the shuffle product. Moreover, this way you can see the coproduct coming from the concatenation of loops and the antipode from reversing orientation (if I remember well).

But here you have to be in a situation when $V^{**}\cong V$.

Let me be a bit more precise. For convenience I will work with $T(V^*)$ equipped with the shuffle product $\star$, the deconcatenation coproduct $\Delta$, and $S$ as you defined it.

Now let me consider the algebra $\mathcal A$ of functionals on $L(V^*)=C_*^0(S^1,V^*)$ (the subscript $*$ means that I ask that $0$ is sent to $0$). There is an algebra monomorphism $T(V^*)\to \mathcal{A}$ given as follows: $$ \xi_1\otimes\cdots\otimes \xi_n\mapsto \left(\gamma\mapsto \int_{0<t_1<\cdots<t_n<1}\xi_1(\gamma(t_1))\dots\xi_n(\gamma(t_n))dt_1\dots dt_n\right) $$

Now observe that the composition of loops and the orientation reversing define algebra morphisms $\Delta_A:\mathcal A\to\mathcal A\otimes\mathcal A$ and $S_A:\mathcal A\to\mathcal A$.

The point is that $\Delta_A$ and $S_A$ do not really satisfy the axioms you want (e.g. coassociativity of $\Delta_A$) BUT their restriction onto the image of $T(V^*)$ does (you have to use an avatar of Stokes' Theorem to see this - or, shortly: iterated integrals are not sensitive to reparametrization), and actually coincide with $\Delta$ and $S$ (very simple computation).

Below are a few algebraic considerations (independant from the above answer).

The main point concerning the antipode is that

any connected filtered bialgebra is a filtered Hopf algebra, the antipode being defined as $S(x)=\sum_{k\geq0}(\eta\circ\epsilon-id)^{*k}(x)$

Here $*$ denotes the convolution product. In your example the bialgebra you consider is actually graded so the result applies. You can find the above claim (and its proof) in these lecture notes (I think you are going to like them) by Dominique Manchon (Corollary II.3.2).

The group algebra of a free group

The degree completion of $T(V)$ is the structure ring of the pro-unipotent completion of a free group.

I hope this can help.

Hi Darij,

A geometric way of thinking about the shuffle algebra

A geometric way of seeing $T(V)$ with the shuffle product is by considering functions on the loop space of $V^*$ (i.e. the space of continuous maps from $S^1$ to $V^*$) that are given by iterated integrals. You'll see that the product of two iterated integrals is precisely the shuffle product. Moreover, this way you can see the coproduct coming from the concatenation of loops and the antipode from reversing orientation (if I remember well).

But here you have to be in a situation when $V^{**}\cong V$.

Let me be a bit more precise. For convenience I will work with $T(V^*)$ equipped with the shuffle product $\star$, the deconcatenation coproduct $\Delta$, and $S$ as you defined it.

Now let me consider the algebra $\mathcal A$ of functionals on $L(V^*)=C_*^0(S^1,V^*)$ (the subscript $*$ means that I ask that $0$ is sent to $0$). There is an algebra monomorphism $T(V^*)\to \mathcal{A}$ given as follows: $$ \xi_1\otimes\cdots \xi_n\mapsto \left(\gamma\mapsto \int_{0<t_1<\cdots<t_n<1}\xi_1(\gamma(t_1))\dots\xi_n(\gamma(t_n))dt_1\dots dt_n\right) $$

Now observe that the composition of loops and the orientation reversing define algebra morphisms $\Delta_A:\mathcal A\to\mathcal A\otimes\mathcal A$ and $S_A:\mathcal A\to\mathcal A$.

The point is that $\Delta_A$ and $S_A$ do not really satisfy the axioms you want (e.g. coassociativity of $\Delta_A$) BUT their restriction onto the image of $T(V^*)$ does (you have to use an avatar of Stokes' Theorem to see this - or, shortly: iterated integrals are not sensitive to reparametrization), and actually coincide with $\Delta$ and $S$ (very simple computation).

The main point concerning the antipode is that

any connected filtered bialgebra is a filtered Hopf algebra, the antipode being defined as $S(x)=\sum_{k\geq0}(\eta\circ\epsilon-id)^{*k}(x)$

Here $*$ denotes the convolution product. In your example the bialgebra you consider is actually graded so the result applies. You can find the above claim (and its proof) in these lecture notes (I think you are going to like them) by Dominique Manchon (Corollary II.3.2).

The group algebra of a free group

The degree completion of $T(V)$ is the structure ring of the pro-unipotent completion of a free group.

I hope this can help.

Hi Darij,

A geometric way of thinking about the shuffle algebra

A geometric way of seeing $T(V)$ with the shuffle product is by considering functions on the loop space of $V^*$ (i.e. the space of continuous maps from $S^1$ to $V^*$) that are given by iterated integrals. You'll see that the product of two iterated integrals is precisely the shuffle product. Moreover, this way you can see the coproduct coming from the concatenation of loops and the antipode from reversing orientation (if I remember well).

But here you have to be in a situation when $V^{**}\cong V$.

Let me be a bit more precise. For convenience I will work with $T(V^*)$ equipped with the shuffle product $\star$, the deconcatenation coproduct $\Delta$, and $S$ as you defined it.

Now let me consider the algebra $\mathcal A$ of functionals on $L(V^*)=C_*^0(S^1,V^*)$ (the subscript $*$ means that I ask that $0$ is sent to $0$). There is an algebra monomorphism $T(V^*)\to \mathcal{A}$ given as follows: $$ \xi_1\otimes\cdots\otimes \xi_n\mapsto \left(\gamma\mapsto \int_{0<t_1<\cdots<t_n<1}\xi_1(\gamma(t_1))\dots\xi_n(\gamma(t_n))dt_1\dots dt_n\right) $$

Now observe that the composition of loops and the orientation reversing define algebra morphisms $\Delta_A:\mathcal A\to\mathcal A\otimes\mathcal A$ and $S_A:\mathcal A\to\mathcal A$.

The point is that $\Delta_A$ and $S_A$ do not really satisfy the axioms you want (e.g. coassociativity of $\Delta_A$) BUT their restriction onto the image of $T(V^*)$ does (you have to use an avatar of Stokes' Theorem to see this - or, shortly: iterated integrals are not sensitive to reparametrization), and actually coincide with $\Delta$ and $S$ (very simple computation).

The main point concerning the antipode is that

any connected filtered bialgebra is a filtered Hopf algebra, the antipode being defined as $S(x)=\sum_{k\geq0}(\eta\circ\epsilon-id)^{*k}(x)$

Here $*$ denotes the convolution product. In your example the bialgebra you consider is actually graded so the result applies. You can find the above claim (and its proof) in these lecture notes (I think you are going to like them) by Dominique Manchon (Corollary II.3.2).

The group algebra of a free group

The degree completion of $T(V)$ is the structure ring of the pro-unipotent completion of a free group.

I hope this can help.

Hi Darij,

the main point concerning the antipode is that

any connected filtered bialgebra is a filtered Hopf algebra, the antipode being defined as $S(x)=\sum_{k\geq0}(\eta\circ\epsilon-id)^{*k}(x)$

Here $*$ denotes the convolution product. In your example the bialgebra you consider is actually graded so the result applies. You can find the above claim (and its proof) in these lecture notes (I think you are going to like them) by Dominique Manchon (Corollary II.3.2).

the group algebra of a free group

The degree completion of $T(V)$ is the structure ring of the pro-unipotent completion of a free group.

geometric way of thinking about the shuffle algebra

A geometric way of seeing $T(V)$ with the shuffle product is by considering functions on the loop space of $V^*$ (i.e. the space of continuous maps from $S^1$ to $V^*$) that are given by iterated integrals. You'll see that the product of two iterated integrals is precisely the shuffle product. Moreover, this way you can see the coproduct coming from the concatenation of loops and the antipode from reversing orientation (if I remember well).

But here you have to be in a situation when $V^{**}\cong V$.

Let me be a bit more precise. For convenience I will work with $T(V^*)$ equipped with the shuffle product $\star$, the deconcatenation coproduct $\Delta$, and $S$ as you defined it.

Now let me consider the algebra $\mathcal A$ of functionals on $L(V^*)=C_*^0(S^1,V^*)$ (the subscript $*$ means that I ask that $0$ is sent to $0$). There is an algebra monomorphism $T(V^*)\to \mathcal{A}$ given as follows: $$ \xi_1\otimes\cdots \xi_n\mapsto (\gamma\mapsto \int_{0<t_1<\cdots<t_n<1}\xi_1(\gamma(t_1))\dots\xi_n(\gamma(t_n))dt_1\dots dt_n $$

Now observe that composition of loops (resp. taking the reverse loop) defines algebra morphism $\Delta_A:\mathcal A\to\mathcal A\otimes\mathcal A$ (resp. antihomomorphism $S_A:\mathcal A\to\mathcal A$).

The point is that $\Delta$ and $S$ do not really satisfy the axioms you want (e.g. coassociativity) BUT their restriction onto the image of $T(V^*)$ does (you have to use an avatar of Stokes' Theorem to see this - or, shortly, iterated integrals are not sensitive to reparametrization), and actually coincide with $\Delta$ and $S$.

I hope this can help.

Hi Darij,

A geometric way of thinking about the shuffle algebra

A geometric way of seeing $T(V)$ with the shuffle product is by considering functions on the loop space of $V^*$ (i.e. the space of continuous maps from $S^1$ to $V^*$) that are given by iterated integrals. You'll see that the product of two iterated integrals is precisely the shuffle product. Moreover, this way you can see the coproduct coming from the concatenation of loops and the antipode from reversing orientation (if I remember well).

But here you have to be in a situation when $V^{**}\cong V$.

Let me be a bit more precise. For convenience I will work with $T(V^*)$ equipped with the shuffle product $\star$, the deconcatenation coproduct $\Delta$, and $S$ as you defined it.

Now let me consider the algebra $\mathcal A$ of functionals on $L(V^*)=C_*^0(S^1,V^*)$ (the subscript $*$ means that I ask that $0$ is sent to $0$). There is an algebra monomorphism $T(V^*)\to \mathcal{A}$ given as follows: $$ \xi_1\otimes\cdots \xi_n\mapsto \left(\gamma\mapsto \int_{0<t_1<\cdots<t_n<1}\xi_1(\gamma(t_1))\dots\xi_n(\gamma(t_n))dt_1\dots dt_n\right) $$

Now observe that the composition of loops and the orientation reversing define algebra morphisms $\Delta_A:\mathcal A\to\mathcal A\otimes\mathcal A$ and $S_A:\mathcal A\to\mathcal A$.

The point is that $\Delta_A$ and $S_A$ do not really satisfy the axioms you want (e.g. coassociativity of $\Delta_A$) BUT their restriction onto the image of $T(V^*)$ does (you have to use an avatar of Stokes' Theorem to see this - or, shortly: iterated integrals are not sensitive to reparametrization), and actually coincide with $\Delta$ and $S$ (very simple computation).

The main point concerning the antipode is that

any connected filtered bialgebra is a filtered Hopf algebra, the antipode being defined as $S(x)=\sum_{k\geq0}(\eta\circ\epsilon-id)^{*k}(x)$

Here $*$ denotes the convolution product. In your example the bialgebra you consider is actually graded so the result applies. You can find the above claim (and its proof) in these lecture notes (I think you are going to like them) by Dominique Manchon (Corollary II.3.2).

The group algebra of a free group

The degree completion of $T(V)$ is the structure ring of the pro-unipotent completion of a free group.

I hope this can help.