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Let $H$ be a Hopf algebra, with invertible antipode, and let $(M,\Delta_M)$ and $(N,\Delta_N)$ be two left $H$-comodules. Now as we all know, we have a left $H$-comodule structure on the tensor product $M \otimes N$, defined by $$ (\Delta_M \otimes \Delta_N)(m \otimes n):= m_{(-1)}n_{(-1)} \otimes m_{(0)} \otimes n_{(0)}, ~~~~~ m \in M, n \in N. $$

Morover, we have canonical right $H$-comodule structures on $M$ and $N$ given by $$ \Delta^R_M(m) := \tau \circ (S \otimes \text{id}) \circ \Delta_M(m) = m_{(0)} \otimes S(m_{(-1)}), $$ and
$$ \Delta^R_N(n) := \tau \circ (S \otimes \text{id}) \circ \Delta_N(n) = n_{(0)} \otimes S(n_{(-1)}), $$ where $\tau$ is the flip operator.

However, it seems to me that since $$ \tau \circ (S \otimes \text{id} \otimes \text{id}) \circ (\Delta_M \otimes \Delta_N)(m \otimes n):= m_{(0)} \otimes n_{(0)} \otimes S(m_{(-1)}n_{(-1)}) $$ while $$ (\Delta_M^R \otimes \Delta_N^R)(m \otimes n):= m_{(0)} \otimes n_{(0)} \otimes S(m_{(-1)})S(n_{(-1)}) = m_{(0)} \otimes n_{(0)} \otimes S(n_{(-1)}m_{(-1)}) $$ we do not in general have $$ \Delta_M^R \otimes \Delta_N^R = (\Delta_M \otimes \Delta_N)^R. $$ Is my reasoning correct here? If it is, then can anyone offer an explanation for why this should happen?

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Unless I am mistaken, what's going on is the following. As vector spaces, there is a canonical isomorphism $M \otimes N \cong N \otimes M$. But this is not generally a map of $H$-comodules. Given left $H$-comodules $M$ and $N$, I will write $\overline M$ and $\overline N$ for the corresponding right $H$-comodules that you have described. Then what you have observed is that $\overline{(\cdot)}$ is an "antihomomorphism" for the tensor product. Namely, under the canonical isomorphism of vector spaces mentioned above, you correctly compute that $\overline M \otimes \overline N \cong \overline{N\otimes M}$ as right $H$-comodules, and not the other way.

Two asides: (1) This is related to various standard facts about $S^{-1}$ and $S^2$ and the various ways to replace the multiplication and/or comultiplication with its opposite map. (2) Nothing is peculiar to comodules in this question, and for many people it is easier to think in terms of modules, so unless you care about finer points like whether infinite-dimensional things are the unions of their finite-dimensional sub-things, you should feel free to try the question with the "co"s dropped.

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