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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 $\mathfrak{so}(3)$?

Novo
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  • Do you mean how is the Lie coalgebra structure defined? – Olivier Bégassat Jul 06 '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 06 '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 06 '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 06 '12 at 13:38

2 Answers2

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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 $$\Theta(H^*)=E^*\otimes F^*-F^*\otimes E^*$$ $$\Theta(F^*)=-2(H^*\otimes F^*-F^*\otimes H^*)$$ $$\Theta(E^*)=2(H^*\otimes E^*-E^*\otimes H^*)$$ The general rule here is that $\Theta (X_k)=\sum_{i,j} r_{i,j,k} X_i\otimes X_j$ where $X_i,X_j$ runs over all pairs of basis elements where $[X_i,X_j]=r_{i,j,k}X_k+\cdots$.

Zbigniew Wojc
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Cheerful Parsnip
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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 06 '12 at 14:51
  • Okay, I will do an example. – Cheerful Parsnip Jul 06 '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 06 '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 $\phi\colon V\to \mathbb R$. – Cheerful Parsnip Jul 06 '12 at 15:20
  • @JimConant: Ah, so the dual of the Lie algebra is just the dual of the vector space. Thanks. – celtschk Jul 06 '12 at 16:09
  • @JimConant Took me some time to fully digest your reply, but I think I understand the relation between the Lie algebra and the Lie coalgebra. Also the fact that it is actually the dual of the vector space is really helpful. – Novo Jul 07 '12 at 15:02
  • @JimConant However I still had one small question. How would one extend the adjoint action on Lie groups to an coadjoint action on the Liecoalgebra? – Novo Jul 07 '12 at 15:07
  • @user29751: you might want to ask that as a separate question. I can tell you how to make sense of the dual of the adjoint representation of the Lie algebra, but when you go to the Lie group level I'm not sure what to do. Maybe there is a Lie cogroup involved, whatever that might be. – Cheerful Parsnip Jul 09 '12 at 03:42
  • @JimConant Ok, thanks. I will do that. – Novo Jul 09 '12 at 09:10
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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.

Vectornaut
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