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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,

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One question mark is enough in most circumstances! –  Mariano Suárez-Alvarez Oct 24 '10 at 14:50

3 Answers 3

up vote 5 down vote accepted

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.

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Indeed, the exercises in Sweedler's book regarding the notation are essentially mandatory for anyone trying to work with these things! –  Mariano Suárez-Alvarez Feb 28 '11 at 16:35

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...)

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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 '10 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 '10 at 17:55
    
if you look at your original definition chris you will see that $\Delta^2 = \Delta$. –  Sean Tilson Oct 24 '10 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 '10 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. –  Loop Space Oct 25 '10 at 7:47

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 $$\Delta^m = (id_C^{\otimes m-1} \otimes \Delta) \circ \Delta^{m-1}.$$

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