6
$\begingroup$

Let $H$ be a complex of vector spaces over some field $k$ which is endowed with the structure of a Hopf algebra object. I have heard several times that if $H$ is concentrated in positive or negative degrees, then the cohomology in degree zero of $H$ is a Hopf algebra in the usual sense. How does one prove such a statement and why is the condition on the degrees important?

$\endgroup$
  • 4
    $\begingroup$ When restricted to these subcategories, the functor $H^0(-)$ is symmetric monoidal and so preserves the Hopf algebra structures. I'm really not sure that this is a research-level question though. $\endgroup$ – Denis Nardin May 6 at 11:30
  • 6
    $\begingroup$ Could you actually explain your answer? $\endgroup$ – hopfology May 6 at 16:28
  • 1
    $\begingroup$ (I've edited the previous comment in response to a flag.) Which part needs explaining? Do you know what a symmetric monoidal ($k$-linear) functor is? Do you understand that if a concept (such as the notion of Hopf algebra object) is definable in the language of symmetric monoidal $k$-linear categories, then a symmetric monoidal $k$-linear functor will map models of that concept in the domain category to ones in the codomain? $\endgroup$ – Todd Trimble Jun 8 at 11:26
2
$\begingroup$

If $C$ is a graded coalgebra (e.g. $C$= the homology of a d.g. Hopf algebra), then $C_0$ is not necesarily a subcoalgebra, because
$$\Delta(C_0)\subset (C\otimes C)_0=\oplus_{n\in\mathbb Z}C_n\otimes C_{-n}$$

For example, $H=k\{x,y\}/(x^2=y^2=xy+yx)$ is a graded Hopf algebra with $|x|=1$ and $|y|=-1$, both $x$ and $y$ primitives (in particular a d.g. Hopf algebra with $d=0$ and agree with its homology).

$H_0=k\oplus kx\wedge y$, and $$ \Delta(x\wedge y)= (x\otimes 1+1\otimes x)(y\otimes 1+1\otimes y)$$ $$= x\wedge y\otimes 1+1\otimes x\wedge y+ x\otimes y-y\otimes x \notin H_0\otimes H_0$$

Of course if $C=\oplus_{n\geq 0} C_n$, then $C_n=0$ for $n<0$ and $C_{-n}=0$ for $n>0$, so, the only nonzero summand in $\oplus_{n\in\mathbb Z}C_n\otimes C_{-n}$ is $C_0\otimes C_0$.

$\endgroup$

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.