1
$\begingroup$

I have a stupid question, but for a Hopf algebra $A$: $A$ and $A\otimes A$ are different Hopf algebras right? As in, being a Hopf algebra simply equips some maps from $A$ and gives us some rules (i.e. three diagrams each for unit, product, counit, coproduct, and extra diagrams for antipode and compatibility). But being a Hopf algebra says little about the algebras $A\otimes^k A$, although these can be given the structure of different Hopf algebras?

I got this confusion from thinking about the universal enveloping algebra, which being $T(\mathfrak{g})/(xy-yx-[x,y])$, say we have $\Delta(x)=x\otimes x$ then it maps from $U(\mathfrak{g})\to U(\mathfrak{g})\otimes U(\mathfrak{g})$ but it is also in $U(\mathfrak{g})$ right?

$\endgroup$
  • $\begingroup$ In general yes, indeed $A$ and $A\otimes A$ often have different dimensions so they cannot be isomorphic. $\endgroup$ – Lord Shark the Unknown May 1 '17 at 6:22
  • $\begingroup$ @LordSharktheUnknown What about the $U(\mathfrak{g})$ case there aswell? $\endgroup$ – Earth Cracks May 1 '17 at 6:42
  • $\begingroup$ Well, $U(\mathfrak g)\otimes U(\mathfrak g)=U(\frak g\oplus\frak g)$. For every Lie theorist's standard example, $\frak g=\mathfrak{sl}(2)$ the centre of $U(\frak g)$ is a polynomial ring in one variable (generated by the Casimir element) so $U(\mathfrak g)\otimes U(\mathfrak g)$'s centre is a polynomial ring of two variables, so $U(\mathfrak g)\otimes U(\mathfrak g)\not\cong U(\mathfrak g)$. $\endgroup$ – Lord Shark the Unknown May 1 '17 at 6:48
  • $\begingroup$ Inside the universal enveloping algebra we have $\Delta(x)=1\otimes x+x\otimes 1$, where $x$ is among the generators of the UEA i.e. the generators are primitives and not grouplikes (as is incorrectly stated at the OP). $\endgroup$ – KonKan May 1 '17 at 19:59
1
$\begingroup$

If I got the point, your problem comes from the fact that the product in $T(\mathfrak{g})$ is given by concatenation, so for $x,y\in\mathfrak{g}$ you may look at $x\otimes y$ both as an element in $T^1(\mathfrak{g})\otimes T^1(\mathfrak{g})$ and in $T^2(\mathfrak{g})$. Formally, you should distinguish the concatenation product $x\cdot y=x\otimes y \in T^2(\mathfrak{g})$ from the "external tensor product" $x\otimes y \in T^1(\mathfrak{g})\otimes T^1(\mathfrak{g})$. What you can say is that the vector spaces $T^1(\mathfrak{g})\otimes T^1(\mathfrak{g})$ and $T^2(\mathfrak{g})$ are isomorphic (exactly via the multiplication) and sometimes you may identify them.

As an example of the importance of distinguish the two tensor products, you may look at the tensor product of two linear maps $f:V\to V'$ and $g:W\to W'$. The writing $f\otimes g$ may mean both the linear map $f\otimes g:V\otimes W\to V'\otimes W'$ and the element $f\otimes g\in \mathsf{Hom}(V,V')\otimes \mathsf{Hom}(W,W')$.

$\endgroup$

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.