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Lately I have been reading about coalgebras and wondering about natural ways one can make any $R$-module into an $R$-coalgebra.

Two examples of such constructions are given by tensor coalgebras and cofree coalgebras. What other examples are there?

In particular, given an $R$-module $M$, do the maps $m\mapsto m\otimes m$ and $m\mapsto m\otimes 1+1\otimes m$ give comultiplications for $M$? If not, what fails?

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The map $m \mapsto m\otimes m$ isn't linear, since for $m_1,m_2\in M$ we have

$$\Delta(m_1 + m_2) = m_1\otimes m_1 + m_1\otimes m_2 + m_2\otimes m_1 + m_2\otimes m_2\neq \Delta(m_1) + \Delta(m_2)$$

So it can't be a coproduct.

Edit: In general $\Delta(m) = m\otimes m$ is not linear for an arbitrary $R$-module $M$, but if $M$ is a free $R$-module then you can define $\Delta$ on the basis of $M$ and extend by linearity. In this case you get a counital and coassociative coalgebra structure on $M$. I'll leave it as an exercise for you to figure out what $\varepsilon$ should be in this case and proving that it and $\Delta$ satisfy the necessary conditions to be counital and coassociative.

On the other hand $m\mapsto m\otimes 1 + 1\otimes m$ is linear, so (assuming you don't require your coalgebras to be counital) that does give a coalgebra structure on $M$. If your definition does require a counit then you'd have to find a map $\varepsilon:M\to R$ such that for each $m\in M$ we have $$ m\otimes \varepsilon(1) + 1\otimes \varepsilon(m) = m\otimes r_1$$ $$ \varepsilon(m)\otimes 1 + \varepsilon(1)\otimes m = r_2\otimes m$$

for some non-zero $r_1, r_2\in R$. The only way to get these tensors to combine into a pure tensor is if for all $m\in M$ we have $\varepsilon(m) = r$ for some $r\in R$. But $\varepsilon$ is linear if and only if $r= 0$, which gives $$ m\otimes 0 + 1\otimes 0 = 0$$ $$ 0\otimes 1 + 0\otimes m = 0$$

Therefore $M$ isn't counital with the coproduct $m\mapsto m\otimes 1 + 1\otimes m$. Furthermore, a coalgebra is coassociative if

$$(\Delta \otimes id)\circ \Delta = (id \otimes \Delta)\circ \Delta $$

If you just check both sides of this equation then you'll see that it's not satisfied. So $m\otimes 1 + 1\otimes m$ isn't coassociative either. So long story short the coalgebra structure that $m\mapsto m\otimes 1 + 1\otimes m$ gives to $M$ is neither counital nor coassociative, which is maybe why you haven't seen it anywhere, because most texts I've seen introduce coalgebras will usually have both of these conditions (or at least the counit condition) as part of the definition.

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  • $\begingroup$ Thank you so much! This is perfect! (though it breaks my heart a bit :P) $\endgroup$
    – Sofia
    Jun 17 at 4:05
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    $\begingroup$ @Sofia Glad I could help! I've edited my answer to include a bit more detail about the first case, which should hopefully give you a bit of relief that it's not a completely lost cause in terms of a simple coalgebra structure :P $\endgroup$
    – SeraPhim
    Jun 22 at 15:55
  • $\begingroup$ Great! I was confused about these two structures because they work for $R[t]$ via $t\mapsto t\otimes t$ and $t\mapsto 1\otimes t+t\otimes 1$, and being free was precisely what was missing. Thank you! $\endgroup$
    – Sofia
    Jul 9 at 11:47

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