3
$\begingroup$

I would like to compute the dual of the hopf algebra $k[\mathbb{Z}/n\mathbb Z]$, where comultiplication is $\Delta(g)=g \otimes g$, coinverse is $S(g)=g^{-1}$ and counit is $\epsilon(g)=[1]_n$. I know the result is the same as the hopf algebra of $k[X]/(X^n-1)$ which corresponds to the group scheme $\mu_n$, but I am not sure how to show this, I have never computed explicitely the dual of anything.

Any hint? Thanks!

$\endgroup$
4
$\begingroup$

For clarity I'll write explicitly the whole group algebra. The Hopf algebra $H = k[\mathbb{Z}/n\mathbb{Z}]$ in question is the $n$-dimensional $k$-vector space with basis $e_0, \ldots, e_{n-1}$, the algebra structure $$\begin{aligned} m: H \otimes H \to H, \quad &e_i \otimes e_j \mapsto e_{i+j} \\ u: k \to H, \quad &1 \mapsto e_0 \end{aligned}$$ the coalgebra structure $$\begin{aligned} \Delta: H \to H \otimes H, \quad &e_i \mapsto e_i \otimes e_i \\ \epsilon: H \to k, \quad &e_i \mapsto 1\end{aligned}$$ and antipode $S: H \to H$, $e_i \mapsto e_{n - i}$ (where subscripts $e_i$ are understood to be modulo $n$).

The dual Hopf algebra $(H^\circ, m^\circ, u^\circ, \Delta^\circ, \epsilon^\circ, S^\circ)$ has underlying vector space $H^\circ = H^*$, so let $f_i: H \to k$ for $i = 1, \ldots, n$ be the basis of $H^*$ dual to the $e_i$. The new unit $u^\circ: k \to H^\circ$ is given by $1 \mapsto \epsilon$, making $\epsilon = f_0 + f_1 + \cdots + f_{n-1}$ the unit element in $H^\circ$.

Then we use the original comultiplication $\Delta$ to figure out the multiplication map $m^\circ$ in the dual:

$$m^\circ(f_i, f_j) = H \xrightarrow{\Delta} H \otimes H \xrightarrow{f_i \otimes f_j} k \otimes k \xrightarrow{\sim} k $$

Checking the action on the basis element $e_k$, $m^\circ(f_i, f_j)(e_k) = f_i(e_k) f_j(e_k)$, and so $m^\circ(f_i, f_i) = f_i$, while $m^\circ(f_i, f_j) = 0$ for $i \neq j$. Notice that $m^\circ(\epsilon, f_i) = f_i$ as expected.

You can go and find the coalgebra structures and antipode on $H^\circ$ yourself, but I'll also comment that the multiplication $m^\circ$ doesn't look obviously like the multiplication in $k[X]/(X^n - 1)$. If $k$ has a primitive $n$th root of unity $\omega$, and the characteristic of $k$ does not divide $n$, then writing

$$ F_i = \frac{1}{n}\left(1 + \omega X + \omega^2 X^2 + \cdots + \omega^{n-1} X^{n-1}\right)$$

makes $F_0, \ldots, F_{n-1}$ a basis of $k[X]/(X^n - 1)$, satisfying $F_i^2 = 1$, $F_i F_j = 0$ for $i \neq j$, and $F_0 + \cdots + F_{n-1} = 1$. (The $F_i$ here are the decomposition of the group algebra ring into its primitive idempotents.) So the algebra structure on $H^\circ$ really does look like $k[X]/(X^n - 1)$, just in a different basis.

$\endgroup$
1
$\begingroup$

Let $H$ be your algebra and let $H^*$ be its dual. Let $g$ be a generator of $Z/nZ$, so that $\{1,g,g^2,\dots,g^{n-1}\}$ is a basis of $H$. Let $\{v_0,\dots,v_{n-1}\}$ be the dual basis of $H^*$, so that $v_i(g^j)=\delta_{i,j}$ for all $i,j\in\{0,\dots,n-1\}$.

The multiplication of $H^*$ is defined as follows: if $a$ and $b$ are two elements of $H^*$,then their product $ab$ in $H^*$ is the composition $a\otimes b\circ\Delta$. As $ab$ is an element of $H^*$, to compute it explicitely it is enough to find its values on the elements of our basis of $H$. If $i\in\{0,\dots,n-\}$ we have $$(ab)(g^i))=(a\otimes b)(\Delta(g^i))=(a\otimes b)(g^i\otimes g^i)=a(g^i)b(g^i).$$ Using this, we find immediately that if $i,j\in\{0,\dots,n-1\}$ we have $$v_iv_j=\delta_{i,j}v_i.$$ This decribes the algebra structure on $H^*$. Can you find its comultplication?

$\endgroup$
  • $\begingroup$ I was computing the multiplication right now and I got that $a \otimes b \circ \Delta$ is the multiplication. Now I am computing the comultiplication. Is it enough to show that if the comultiplication and multiplication are the same then the Hopf algebra are the same? Or am I to compute also inverse, coinverse, unit and counit? $\endgroup$ – Ale Sep 28 '17 at 5:30
  • $\begingroup$ Also, I really don't get what what I can infer from what you wrote, the multiplication in $\mu_n$ does not satisfy $e_ie_j=\delta_{i,j}e_i$ if I am not mistaken. It should just be: $m(X^i \otimes X_j)=X^{i+j}$. $\endgroup$ – Ale Sep 28 '17 at 5:51
  • $\begingroup$ Well, my computation is correct. Joppy's answre, which is exactly the same as mine, gets the same multiplication. $\endgroup$ – Mariano Suárez-Álvarez Sep 28 '17 at 14:46

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.