Welcome to MathOverflow

Question

0

3

Let $C$ be a coalgebra and $\Delta: C \to C\otimes C$ a co-multiplication map. Then, due the co-associative property we can consider $\Delta^m$. But how is defined $\Delta^{m}: C \to C^{\otimes m}$???

Given $f,g \in C$ and $1\leq k \leq m$ can we have \begin{align*}\Delta^{m-1}(fg)&=\Delta^{k-1}(fg) \otimes id^{m-k} + \Delta^{k-1}(f)\otimes \Delta^{m-k-1}(g) \\ &+\Delta^{k-1}(g)\otimes \Delta^{m-k-1}(f)+id^{\otimes k} \otimes \Delta^{m-k-1}(fg)???\end{align*}

Thanks,

edited Feb 28 2011 at 13:47

Matthew Daws
5,489●1●8●28

asked Oct 24 2010 at 14:16

Chris
339●7

11

One question mark is enough in most circumstances! – Mariano Suárez-Alvarez Oct 24 2010 at 14:50

Konstantin Ardakov · Accepted Answer · 2011-02-28 11:46:47Z UTC

To add to the answer above, I'd like to advertise the so-called sumless Sweedler notation here.

This notation works as follows: let $C$ be a coalgebra; then if $c \in C$ then we write $\Delta(c) = c_1 \otimes c_2$ as an abbreviation of the more precise $\Delta(c) = \sum_{i=1}^k c_{i1} \otimes c_{i2}$ for some $c_{i1}, c_{i2} \in C, i = 1,\ldots, k$. Then if we wish to consider $(1 \otimes \Delta) \circ \Delta : C \to C\otimes C \otimes C$ we simply write

$$[(1 \otimes \Delta) \circ \Delta] (c) = c_1 \otimes c_2 \otimes c_3.$$

In this notation, the coassociativity axiom $(1 \otimes \Delta) \circ \Delta = (\Delta \otimes 1) \circ \Delta$ becomes the inevitable

$$ c_1 \otimes (c_2 \otimes c_3) = (c_1 \otimes c_2) \otimes c_3.$$

Sweedler's book "Hopf algebras", Susan Montgomery's book "Hopf algebras and their actions on rings" and the Wikipedia article http://en.wikipedia.org/wiki/Coalgebra are good references.

Mariano Suárez-Alvarez · Answer 2 · 2010-10-24 14:43:27Z UTC

7

We usually define $$\Delta^3=(\mathrm{id}_C\otimes\Delta)\circ\Delta,$$ and, more generally, $$\Delta^{k+1}=(\mathrm{id}_C^{k-1}\otimes\Delta)\circ\Delta^k.$$

(Your last formula does not make sense in a coalgebra: you seem to be multiplying elements of $C$; even in bialgebra, though, things like "$\Delta(fg)\otimes\mathrm{id}$" are pretty strange...)

edited Oct 24 2010 at 14:43

answered Oct 24 2010 at 14:35

Mariano Suárez-Alvarez
30.5k●7●68●153

	Both equations are not consistent! Anyway, have you meant $\Delta^2$ in the first formula or the power is wrong in the second one? – Chris Oct 24 2010 at 17:34
	One more point. Observe that the usual action of $fg$ on $u\otimes v$, where $f,g$ are two elements in the Universal Enveloping Algebra $U(G)$ of a Lie algebra $G$ and $u,v$ are elements of a representation $V$ of $G$, is given by $fg (u \otimes v)=fgu \otimes v + fu \otimes gv + gu\otimes fv + u\otimes fgv$ using the comultiplication, right? How to state this fact for $V^{\otimes n}$, i.e. $fg$ acting on $u \otimes v$, where $u=\otimes_{i=1}^{n-k} u_i$ and $v=\otimes_{i=1}^k v_i$? Thanks, – Chris Oct 24 2010 at 17:55
	if you look at your original definition chris you will see that $\Delta^2 = \Delta$. – Sean Tilson Oct 24 2010 at 17:59
	@Chris: The numbering convention is for $\Delta^n$ to denote the map $C \to C^{\otimes n}$ --- it has precisely $n$ outputs. You can extend this backwards: $\Delta^2 = \Delta$, $\Delta^1 = \operatorname{id}$, and, if your coalgebra is counital, $\Delta^0$ is the counit. – Theo Johnson-Freyd Oct 24 2010 at 19:00
	And in a bialgebra then the comultiplication is an algebra homomorphism so one would have $\Delta(f g) = \Delta(f) \Delta(g)$ and so on for the iterates. – Andrew Stacey Oct 25 2010 at 7:47

Ralph · Answer 3 · 2011-02-28 22:00:54Z UTC

When working with coalgebras I often find it helpful to dualize and to consider the corresponding situation for multiplication $\mu: A \otimes A \to A$ with an algebra $A$.

In this point of view $\Delta^m$ corresponds to $\mu^m: A^{m+1} \to A$, given by $$\mu^m(a_0 \otimes ... \otimes a_m) = (a_0 ... a_{m-1}) a_m.$$ It's easy to see that this formula is equivalent to $$\mu^m = \mu \circ (\mu^{m-1} \otimes id_A).$$ Dualizing again yields $$\Delta^m = (\Delta^{m-1} \otimes id_C) \circ \Delta.$$ BTW: If $\Delta$ is coassociative this equals Mariano's formula $$\Delta^m = (id_C^{\otimes m-1} \otimes \Delta) \circ \Delta^{m-1}.$$

How to work with co-multiplication?

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

3 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.