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I don't understand how the Lie coalgebra is defined. The literature is never really explicit in how it is constructed. So I was wondering if anybody could supply me with a simple example of how the Lie coalgebra is constructed. Let's say for $\mathfrak{so}(3)$?

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Do you mean how is the Lie coalgebra structure defined? –  Olivier Bégassat Jul 6 '12 at 13:00
    
Can you cite an occurrence of your "dual of a Lie algebra". My first thought would be a Lie algebra where the Lie bracket is defined with its arguments reversed (although this would be more properly be called the oppositive Lie algebra), but I might guess wrong. –  Marc van Leeuwen Jul 6 '12 at 13:35
    
@OlivierBégassat Yes, I mean the Lie coalgebra, the dual structure on the Lie algebra. I will change this in the post. –  Novo Jul 6 '12 at 13:35
    
@MarcvanLeeuwen The dual structe of a Lie algebra is terminology wise a bit unclear. I should call it Lie coalgebra –  Novo Jul 6 '12 at 13:38

2 Answers 2

up vote 6 down vote accepted

Here is one way of defining a Lie coalgebra structure on a vector space $V$. The cobracket is a map $\Theta\colon V\to V\otimes V$ satisfying antisymmetry and co-Jacobi. Antisymmetry means that $\Theta$ induces a map $V\to V\wedge V$. Then co-Jacobi means that if $\Theta(v)=\sum_{i\in I} r_i v_i\wedge w_i$ where $r_i\in\mathbb R$, then $\sum_{i\in I} r_i \Theta(v_i)\wedge w_i + v_i\wedge \Theta(w_i)=0$. One way to think of this is that we have extended $\Theta$ to the entire exterior algebra $\Lambda V$ as a derivation, and the co-Jacobi identity is equivalent to saying that $\Theta^2=0$.

This is dual to one formulation of the Lie algebra axioms. Given a map $b:V\wedge V\to V$, extend it to $\Lambda V$ as a coderivation. Then the Jacobi identity is equivalent to $b^2=0$.

The connection between a Lie algebra $\mathfrak g$ and its dual $\mathfrak g^*$ is that the bracket $b\colon \mathfrak g\otimes \mathfrak g\to \mathfrak g$ is dual to $\Theta\colon \mathfrak g^*\to\mathfrak g^*\otimes \mathfrak g^*$. For example, take $\mathfrak{sl}_2$, with three generators, $E,F,H$ with bracket defined by $$[H,E]=2E,\,\,\, [H,F]=-2F,\,\,\, [E,F]=H.$$

Now take a basis for $\mathfrak{sl}_2^*$ to be $E^*,F^*,H^*$, the dual generators to the basis $E,F,H$. Then we have $$\Theta(H^*)=E^*\otimes F^*-F^*\otimes E^*$$ $$\Theta(F^*)=-\frac{1}{2}(H^*\otimes F^*-F^*\otimes H^*)$$ $$\Theta(E^*)=\frac{1}{2}(H^*\otimes E^*-E^*\otimes H^*)$$ The general rule here is that $\Theta (X)=\sum_{i,j} r_{i,j} X_i\otimes X_j$ where $X_i,X_j$ runs over all pairs of basis elements where $[X_i,X_j]=cX+\cdots$ with $c\neq 0$ and in that case $r_{i,j}:=c^{-1}$.

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It is still unclear to me what the connection between a Lie algebra and its Lie coalgebra is. More specifically given a Lie algebra how can I determine what the associated Lie coalgebra is? Could you illustrate this with a simple example? –  Novo Jul 6 '12 at 14:51
    
Okay, I will do an example. –  Grumpy Parsnip Jul 6 '12 at 14:53
    
Assuming that $H=\left(\begin{matrix}1&0\\0&-1\end{matrix}\right)$, $E=\left(\begin{matrix}0&1\\0&0\end{matrix}\right)$, $F=\left(\begin{matrix}0&0\\1&0\end{matrix}\right)$, then what are $H^*$, $E^*$ and $F^*$? –  celtschk Jul 6 '12 at 15:14
    
$H^*$ is the functional that evaluates to $1$ on $H$ and $0$ on $E$ and $F$. Recall that the dual of a vector space is the set of all homomorphisms $\phi\colon V\to \mathbb R$. –  Grumpy Parsnip Jul 6 '12 at 15:20
    
@JimConant: Ah, so the dual of the Lie algebra is just the dual of the vector space. Thanks. –  celtschk Jul 6 '12 at 16:09

As far as I can tell, the notion of a Lie coalgebra was first introduced in the paper "Lie coalgebras" by Walter Michaelis (Advances in Mathematics, Volume 38, Issue 1). Although I haven't compared them carefully, I think the definition there matches the one Grumpy Parsnip gave.

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