Alternative Cotensor Definition

Question

Let $H$ be a Hopf algebra, and $(M,\rho_r)$ and $(N,\rho_l)$ right and left $H$-comodules respectively. As usual, we define their cotensor product to be $$ M \square_H N := \text{ker}\{(\rho_r \otimes \text{id} - \text{id} \otimes \rho_l ): M \otimes N \to M \otimes H \otimes N\}. $$
I have seen an alternative version of the definition $$ M \square_H N := \{\sum_i m_i \otimes n_i, \sum (m_i)_0 \otimes (n_i)_0 \otimes (m_i)_1 (S(n_i)_{-1}) = \sum_i m_i \otimes n_i \otimes 1 \}. $$ Are they equivalent, and if so, then how does one show this?

Answer 1

Yes, they are equivalent.

Let me restate this fact in more proof-friendly terms:

Let $\mathbf{k}$ be a commutative ring. All tensor signs $\otimes$ will be understood to mean $\otimes_{\mathbf{k}}$. Let $H$ be a $\mathbf{k}$-Hopf algebra with antipode $S:H\rightarrow H$. Let $\left( M,\rho_{r}\right) $ be a right $H$-module (so that $\rho_{r}:M\rightarrow M\otimes H$ is a $\mathbf{k}$-linear map satisfying certain axioms). Let $\left( N,\rho_{\ell }\right) $ be a left $H$-module (so that $\rho_{\ell}:N\rightarrow H\otimes N$ is a $\mathbf{k}$-linear map satisfying certain axioms). We shall use the sum-free Sweedler notation: e.g., we write $h_{\left( 1\right) }\otimes h_{\left( 2\right) }$ for the image of a $h\in H$ under the comultiplication of $H$; we will write $m_{\left( 0\right) }\otimes m_{\left( 1\right) }$ for $\rho_{r}\left( m\right) $ when $m\in M$; and we will write $n_{\left( -1\right) }\otimes n_{\left( 0\right) }$ for $\rho_{\ell}\left( n\right) $ when $n\in N$.

Define a $\mathbf{k}$-linear map $\Psi:M\otimes N\rightarrow M\otimes H\otimes N$ by the requirement that

$\Psi\left( m\otimes n\right) =m_{\left( 0\right) }\otimes m_{\left( 1\right) }S\left( n_{\left( -1\right) }\right) \otimes n_{\left( 0\right) }$ for all $m\in M$ and $n\in N$.

Define a $\mathbf{k}$-linear map $\Omega:M\otimes N\rightarrow M\otimes H\otimes N$ by the requirement that

$\Omega\left( m\otimes n\right) =m\otimes1\otimes n$ for all $m\in M$ and $n\in N$.

Let $\tau_{N}:H\otimes N\rightarrow N\otimes H$ be the twist map (i.e., the $\mathbf{k}$-linear map sending each $h\otimes n$ to $n\otimes h$)

Your second definition of $M\square_{H}N$ then rewrites as follows:

$M\square_{H}N=\ker\left( \left( \operatorname*{id}\nolimits_{M}\otimes \tau\right) \circ\left( \Psi-\Omega\right) \right) $

(I will explain more carefully below why this is equivalent to your second definition).

Since $\operatorname*{id}\nolimits_{M}\otimes\tau$ is an isomorphism of $\mathbf{k}$-modules, this is equivalent to

(1) $M\square_{H}N=\ker\left( \Psi-\Omega\right) $.

So we must prove (1).

Let $\mu$ denote the multiplication map $H\otimes H\rightarrow H$ of the $\mathbf{k}$-algebra $H$. Let $\Delta$ and $\varepsilon$ denote the comultiplication and the counity of the $\mathbf{k}$-coalgebra $H$. We claim that

(2) $\left( \operatorname*{id}\nolimits_{M}\otimes\mu\otimes \operatorname*{id}\nolimits_{N}\right) \circ\left( \operatorname*{id} \nolimits_{M\otimes H}\otimes\rho_{\ell}\right) \circ\Psi=\rho_{r} \otimes\operatorname*{id}\nolimits_{N}$

as $\mathbf{k}$-linear maps $M\otimes N\rightarrow M\otimes H\otimes N$. Indeed, for any $m\in M$ and $n\in N$, we have

$\left( \left( \operatorname*{id}\nolimits_{M}\otimes\mu\otimes \operatorname*{id}\nolimits_{N}\right) \circ\left( \operatorname*{id} \nolimits_{M\otimes H}\otimes\rho_{\ell}\right) \circ\Psi\right) \left( m\otimes n\right) $

$=\left( \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id} \nolimits_{N}\right) \left( \left( \operatorname*{id}\nolimits_{M\otimes H}\otimes\rho_{\ell}\right) \left( \underbrace{\Psi\left( m\otimes n\right) }_{=m_{\left( 0\right) }\otimes m_{\left( 1\right) }S\left( n_{\left( -1\right) }\right) \otimes n_{\left( 0\right) }}\right) \right) $

$=\left( \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id} \nolimits_{N}\right) \left( \underbrace{\left( \operatorname*{id} \nolimits_{M\otimes H}\otimes\rho_{\ell}\right) \left( m_{\left( 0\right) }\otimes m_{\left( 1\right) }S\left( n_{\left( -1\right) }\right) \otimes n_{\left( 0\right) }\right) }_{=m_{\left( 0\right) }\otimes m_{\left( 1\right) }S\left( n_{\left( -1\right) }\right) \otimes \rho_{\ell}\left( n_{\left( 0\right) }\right) }\right) $

$=\left( \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id} \nolimits_{N}\right) \left( m_{\left( 0\right) }\otimes m_{\left( 1\right) }S\left( n_{\left( -1\right) }\right) \otimes\underbrace{\rho _{\ell}\left( n_{\left( 0\right) }\right) }_{=\left( n_{\left( 0\right) }\right) _{\left( -1\right) }\otimes\left( n_{\left( 0\right) }\right) _{\left( 0\right) }}\right) $

$=\left( \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id} \nolimits_{N}\right) \left( m_{\left( 0\right) }\otimes m_{\left( 1\right) }S\left( n_{\left( -1\right) }\right) \otimes\left( n_{\left( 0\right) }\right) _{\left( -1\right) }\otimes\left( n_{\left( 0\right) }\right) _{\left( 0\right) }\right) $

$=\left( \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id} \nolimits_{N}\right) \left( m_{\left( 0\right) }\otimes m_{\left( 1\right) }S\left( n_{\left( -2\right) }\right) \otimes n_{\left( -1\right) }\otimes n_{\left( 0\right) }\right) $

$=m_{\left( 0\right) }\otimes m_{\left( 1\right) }\underbrace{S\left( n_{\left( -2\right) }\right) n_{\left( -1\right) }}_{=\varepsilon\left( n_{\left( -1\right) }\right) 1}\otimes n_{\left( 0\right) }$

$=m_{\left( 0\right) }\otimes m_{\left( 1\right) }\varepsilon\left( n_{\left( -1\right) }\right) 1\otimes n_{\left( 0\right) } =\underbrace{m_{\left( 0\right) }\otimes m_{\left( 1\right) }}_{=\rho _{r}\left( m\right) }\otimes\underbrace{\varepsilon\left( n_{\left( -1\right) }\right) n_{\left( 0\right) }}_{=n}$

$=\rho_{r}\left( m\right) \otimes n=\left( \rho_{r}\otimes \operatorname*{id}\nolimits_{N}\right) \left( m\otimes n\right) $.

This proves (2). Next, we claim that

(3) $\left( \operatorname*{id}\nolimits_{M}\otimes\mu\otimes \operatorname*{id}\nolimits_{N}\right) \circ\left( \operatorname*{id} \nolimits_{M\otimes H}\otimes\rho_{\ell}\right) \circ\Omega =\operatorname*{id}\nolimits_{M}\otimes\rho_{\ell}$

as $\mathbf{k}$-linear maps $M\otimes N\rightarrow M\otimes H\otimes N$. Indeed, for any $m\in M$ and $n\in N$, we have

$\left( \left( \operatorname*{id}\nolimits_{M}\otimes\mu\otimes \operatorname*{id}\nolimits_{N}\right) \circ\left( \operatorname*{id} \nolimits_{M\otimes H}\otimes\rho_{\ell}\right) \circ\Omega\right) \left( m\otimes n\right) $

$=\left( \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id} \nolimits_{N}\right) \left( \left( \operatorname*{id}\nolimits_{M\otimes H}\otimes\rho_{\ell}\right) \left( \underbrace{\Omega\left( m\otimes n\right) }_{=m\otimes1\otimes n}\right) \right) $

$=\left( \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id} \nolimits_{N}\right) \left( \underbrace{\left( \operatorname*{id} \nolimits_{M\otimes H}\otimes\rho_{\ell}\right) \left( m\otimes1\otimes n\right) }_{=m\otimes1\otimes\rho_{\ell}\left( n\right) }\right) $

$=\left( \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id} \nolimits_{N}\right) \left( m\otimes1\otimes\underbrace{\rho_{\ell}\left( n\right) }_{=n_{\left( -1\right) }\otimes n_{\left( 0\right) }}\right) $

$=\left( \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id} \nolimits_{N}\right) \left( m\otimes1\otimes n_{\left( -1\right) }\otimes n_{\left( 0\right) }\right) $

$=m\otimes\underbrace{1n_{\left( -1\right) }\otimes n_{\left( 0\right) } }_{=n_{\left( -1\right) }\otimes n_{\left( 0\right) }=\rho_{\ell}\left( n\right) }=m\otimes\rho_{\ell}\left( n\right) $

$=\left( \operatorname*{id}\nolimits_{M}\otimes\rho_{\ell}\right) \left( m\otimes n\right) $.

This proves (3). Subtracting (3) from (2), we obtain

$\left( \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id} \nolimits_{N}\right) \circ\left( \operatorname*{id}\nolimits_{M\otimes H}\otimes\rho_{\ell}\right) \circ\left( \Psi-\Omega\right) =\rho_{r} \otimes\operatorname*{id}\nolimits_{N}-\operatorname*{id}\nolimits_{M} \otimes\rho_{\ell}$.

Hence,

(4) $\ker\left( \Psi-\Omega\right) \subseteq\ker\left( \rho_{r} \otimes\operatorname*{id}\nolimits_{N}-\operatorname*{id}\nolimits_{M} \otimes\rho_{\ell}\right) $.

On the other hand, let $\gamma:H\otimes N\rightarrow H\otimes N$ be the $\mathbf{k}$-linear map defined by the requirement that

$\gamma\left( h\otimes n\right) =hS\left( n_{\left( -1\right) }\right) \otimes n_{\left( 0\right) }$ for all $h\in H$ and $n\in N$.

(Avoiding Sweedler notation, we could define $\gamma$ as $\left( \mu \otimes\operatorname*{id}\nolimits_{N}\right) \circ\left( \operatorname*{id} \nolimits_{H}\otimes S\otimes\operatorname*{id}\nolimits_{N}\right) \circ\left( \operatorname*{id}\nolimits_{H}\otimes\rho_{\ell}\right) $, but we wouldn't gain much from this definition.)

Now, I claim that

(5) $\left( \operatorname*{id}\nolimits_{M}\otimes\gamma\right) \circ\left( \rho_{r}\otimes\operatorname*{id}\nolimits_{N}\right) =\Psi$

as $\mathbf{k}$-linear maps $M\otimes N\rightarrow M\otimes H\otimes N$. Indeed, for any $m\in M$ and $n\in N$, we have

$\left( \left( \operatorname*{id}\nolimits_{M}\otimes\gamma\right) \circ\left( \rho_{r}\otimes\operatorname*{id}\nolimits_{N}\right) \right) \left( m\otimes n\right) $

$=\left( \operatorname*{id}\nolimits_{M}\otimes\gamma\right) \left( \underbrace{\left( \rho_{r}\otimes\operatorname*{id}\nolimits_{N}\right) \left( m\otimes n\right) }_{=\rho_{r}\left( m\right) \otimes n}\right) =\left( \operatorname*{id}\nolimits_{M}\otimes\gamma\right) \left( \underbrace{\rho_{r}\left( m\right) \otimes n}_{=m_{\left( 0\right) }\otimes m_{\left( 1\right) }\otimes n}\right) $

$=\left( \operatorname*{id}\nolimits_{M}\otimes\gamma\right) \left( m_{\left( 0\right) }\otimes m_{\left( 1\right) }\otimes n\right) =m_{\left( 0\right) }\otimes\underbrace{\gamma\left( m_{\left( 1\right) }\otimes n\right) }_{\substack{=m_{\left( 1\right) }S\left( n_{\left( -1\right) }\right) \otimes n_{\left( 0\right) }\\\text{(by the definition of }\gamma\text{)}}}$

$=m_{\left( 0\right) }\otimes m_{\left( 1\right) }S\left( n_{\left( -1\right) }\right) \otimes n_{\left( 0\right) }=\Psi\left( m\otimes n\right) $.

This proves (5). Furthermore, I claim that

(6) $\left( \operatorname*{id}\nolimits_{M}\otimes\gamma\right) \circ\left( \operatorname*{id}\nolimits_{M}\otimes\rho_{\ell}\right) =\Omega$

as $\mathbf{k}$-linear maps $M\otimes N\rightarrow M\otimes H\otimes N$. Indeed, for any $m\in M$ and $n\in N$, we have

$\left( \left( \operatorname*{id}\nolimits_{M}\otimes\gamma\right) \circ\left( \operatorname*{id}\nolimits_{M}\otimes\rho_{\ell}\right) \right) \left( m\otimes n\right) $

$=\left( \operatorname*{id}\nolimits_{M}\otimes\gamma\right) \left( \underbrace{\left( \operatorname*{id}\nolimits_{M}\otimes\rho_{\ell}\right) \left( m\otimes n\right) }_{=m\otimes\rho_{\ell}\left( n\right) }\right) =\left( \operatorname*{id}\nolimits_{M}\otimes\gamma\right) \left( \underbrace{m\otimes\rho_{\ell}\left( n\right) }_{=m\otimes n_{\left( -1\right) }\otimes n_{\left( 0\right) }}\right) $

$=\left( \operatorname*{id}\nolimits_{M}\otimes\gamma\right) \left( m\otimes n_{\left( -1\right) }\otimes n_{\left( 0\right) }\right) =m\otimes\underbrace{\gamma\left( n_{\left( -1\right) }\otimes n_{\left( 0\right) }\right) }_{\substack{=n_{\left( -1\right) }S\left( \left( n_{\left( 0\right) }\right) _{\left( -1\right) }\right) \otimes\left( n_{\left( 0\right) }\right) _{\left( 0\right) }\\\text{(by the definition of }\gamma\text{)}}}$

$=m\otimes n_{\left( -1\right) }S\left( \left( n_{\left( 0\right) }\right) _{\left( -1\right) }\right) \otimes\left( n_{\left( 0\right) }\right) _{\left( 0\right) }=m\otimes\underbrace{n_{\left( -2\right) }S\left( n_{\left( -1\right) }\right) }_{=\varepsilon\left( n_{\left( -1\right) }\right) 1}\otimes n_{\left( 0\right) }$

$=m\otimes\varepsilon\left( n_{\left( -1\right) }\right) 1\otimes n_{\left( 0\right) }=m\otimes1\otimes\underbrace{\varepsilon\left( n_{\left( -1\right) }\right) n_{\left( 0\right) }}_{=n}=m\otimes1\otimes n=\Omega\left( n\right) $.

This proves (6). Subtracting (6) from (5), we obtain

(7) $\left( \operatorname*{id}\nolimits_{M}\otimes\gamma\right) \circ\left( \rho_{r}\otimes\operatorname*{id}\nolimits_{N}-\operatorname*{id} \nolimits_{M}\otimes\rho_{\ell}\right) =\Psi-\Omega$.

Hence,

$\ker\left( \rho_{r}\otimes\operatorname*{id}\nolimits_{N}-\operatorname*{id} \nolimits_{M}\otimes\rho_{\ell}\right) \subseteq\ker\left( \Psi -\Omega\right) $.

Combining this with (4), we find

$\ker\left( \rho_{r}\otimes\operatorname*{id}\nolimits_{N}-\operatorname*{id} \nolimits_{M}\otimes\rho_{\ell}\right) =\ker\left( \Psi-\Omega\right) $.

Now, the first definition of $M\square_{H}N$ yields

$M\square_{H}N=\ker\left( \rho_{r}\otimes\operatorname*{id}\nolimits_{N} -\operatorname*{id}\nolimits_{M}\otimes\rho_{\ell}\right) =\ker\left( \Psi-\Omega\right) $

$=\ker\left( \left( \operatorname*{id}\nolimits_{M}\otimes\tau\right) \circ\left( \Psi-\Omega\right) \right) $

(since $\operatorname*{id}\nolimits_{M}\otimes\tau$ is a $\mathbf{k}$-module isomorphism (since $\tau$ is a $\mathbf{k}$-module somorphism))

(8) $=\left\{ t\in M\otimes N\ \mid\ \left( \left( \operatorname*{id} \nolimits_{M}\otimes\tau\right) \circ\left( \Psi-\Omega\right) \right) \left( t\right) =0\right\} $.

But if $t\in M\otimes N$ is given as a sum of pure tensors as follows:

$t=\sum_{i}m_{i}\otimes n_{i}$,

then

$\left( \left( \operatorname*{id}\nolimits_{M}\otimes\tau\right) \circ\left( \Psi-\Omega\right) \right) \left( t\right) $

$=\sum_{i}\left( m_{i}\right) _{\left( 0\right) }\otimes\left( n_{i}\right) _{\left( 0\right) }\otimes\left( m_{i}\right) _{\left( -1\right) }S\left( \left( n_{i}\right) _{\left( 1\right) }\right) -\sum_{i}m_{i}\otimes n_{i}\otimes1$,

and therefore (8) rewrites as follows:

$M\square_{H}N$

$=\left\{ \sum_{i}m_{i}\otimes n_{i}\in M\otimes N\ \mid\ \sum_{i}\left( m_{i}\right) _{\left( 0\right) }\otimes\left( n_{i}\right) _{\left( 0\right) }\otimes\left( m_{i}\right) _{\left( -1\right) }S\left( \left( n_{i}\right) _{\left( 1\right) }\right) =\sum_{i}m_{i}\otimes n_{i} \otimes1\right\} $.