6
$\begingroup$

Let $\mathbf{NCoalg}$ be the category of non-unital coassociative conilpotent coalgebras (where "conilpotent" means that for any element there exists a power of comultiplication that vanishes on this element). We have a forgetful functor $\mathbf{NCoalg} \to \mathbf{Vect}$. Its seems that the right adjoint has to be the functor of non-unital tensor coalgebra: $$T(V)=V\oplus V^{\otimes 2}\oplus V^{\otimes3}\oplus \cdots ,$$ $$ \Delta(v_1\otimes v_2\otimes \cdots \otimes v_n) = (v_1)\otimes(v_2\otimes\cdots\otimes v_n)+\cdots+(v_1 \otimes\cdots\otimes v_{n-1})\otimes(v_n).$$

Indeed, let $C$ be a coalgebra and $V$ a vector space. Consider a linear map $f: C \to V$. We would like to define a coalgebra map $\tilde{f}:C\to T(V)$ as follows: $$ c \mapsto f(c)+f^{\otimes2}\Delta(c)+f^{\otimes3}\Delta^2(c)+\cdots. $$

The sum is in fact finite because $C$ is conilpotent.

However, how can I show that $\tilde{f}^{\otimes2}\Delta(c)=\Delta(\tilde{f}(c))$? I guess it can be done with some abominable coordinates but they appeared too abominable and I couldn't break through. Is there a better way (more categorical perhaps)?

$\endgroup$
  • $\begingroup$ If $v\in V\subset T(V)$, what would $\Delta(v)$ be according to this formula ? Is it zero ? $\endgroup$ – Captain Lama Apr 18 '16 at 11:02
  • 1
    $\begingroup$ @CaptainLama Yes, it is zero ($v$ is primitive). $\endgroup$ – Najib Idrissi Apr 18 '16 at 11:05
1
$\begingroup$

It's actually not that hard, but it needs a little care (for regular people like me who find working with coalgebras counter-intuitive :) ).

It all boils down to the fact that if $\Delta(c) = \sum_i x_i\otimes y_i$ then $$\Delta\left(f^{\otimes (n+2)}\Delta^{n+1}(c)\right) = \sum_i \sum_{p+q=n} \left(f^{\otimes (p+1)}\Delta^p(x_i)\right)\otimes \left(f^{\otimes (q+1)}\Delta^q(y_i)\right).$$

Indeed, coassociativity tells you that in $C^{\otimes (n+2)}$, $\Delta^{n+1}(c) = \sum_i \Delta^p(x_i)\otimes \Delta^q(y_i)$ for any $p,q$ such that $p+q = n$, when you identify $C^{\otimes (n+2)} \simeq C^{\otimes (p+1)}\otimes C^{\otimes (q+1)}$.

So if we call $\Phi_{p,q}: V^{\otimes (n+2)} \to V^{\otimes (p+1)}\otimes V^{\otimes (q+1)}$ the canonical isomorphism you get $\Phi_{p,q}\left(f^{\otimes (n+2)}\Delta^{n+1}(c)\right) = \sum_i \left(f^{\otimes (p+1)}\Delta^p(x_i)\right)\otimes \left( f^{\otimes (q+1)}\Delta^q(y_i)\right)$.

But now if $x\in V^{\otimes (n+2)}$ then $\Delta(x) = \sum_{p+q=n} \Phi_{p,q}(x)$ by definition, so if you apply $f^{\otimes(n+2)}$ and $\Delta$ to the formula above expressing the coassociativity, you get the formula emphasized at the beginning.

Now we just need to unravel definitions :

$$\begin{eqnarray*} \widetilde{f}^{\otimes 2}\Delta(c) & = & \sum_i \widetilde{f}(x_i)\otimes \widetilde{f}(y_i) \\ & = & \sum_i \left( \sum_{p\geqslant 0} f^{\otimes (p+1)}\Delta^p(x_i)\right)\otimes \left( \sum_{q\geqslant 0} f^{\otimes (q+1)}\Delta^q(y_i)\right) \\ & = & \sum_{n\geqslant 0} \sum_i \sum_{p+q=n} \left(f^{\otimes (p+1)}\Delta^p(x_i)\right)\otimes \left(f^{\otimes (q+1)}\Delta^q(y_i)\right) \\ & = & \sum_{n\geqslant 0} \Delta\left(f^{\otimes (n+2)}\Delta^{n+1}(c)\right) \\ & = & \sum_{n\geqslant 0} \Delta\left(f^{\otimes (n+1)}\Delta^n(c)\right) \\ & = & \Delta(\widetilde{f}(c)). \end{eqnarray*}$$

$\endgroup$
  • $\begingroup$ Thanks a lot! I'm indeed a very regular person who finds multiplication a lot handier than comultiplication. But yes, your explanation was very clear. $\endgroup$ – Polydarya Apr 18 '16 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.