I'm having this lecture on QM and we are giving an introduction on Lie Groups.

So... this week we have been talking about central extensions of LG (such as Galilean) and related to this popped up the 2-cocycles. All I know is that they should relate somehow the phases of its projective representation and by that should have a concrete property that defines them.

  • $\begingroup$ en.wikipedia.org/wiki/… In particular, "The set of isomorphism classes of central extensions of G by A is in one-to-one correspondence with the cohomology group H2(G,A), where the action of G on A is trivial." $\endgroup$ Oct 24, 2011 at 17:45

1 Answer 1


If you have a projective representation of a Lie group $G$, you can directly construct a central extension in the following way: Try to make the projective representation into an honest representation, and keep track of how it fails.

More specifically, a projective representation $\rho$ attaches to each element $g$ a whole line's worth of linear transformations, i.e., a collection of matrices of the form $\lambda A$ for some matrix $A$, and $\lambda$ complex. To make this into an honest representation, you choose for each $g$ a nonzero linear transformation $\hat{\rho}(g)$ from the line $\rho(g)$. The reason this might fail is that $\hat{\rho}(g) \hat{\rho}(h)$ may not equal $\hat{\rho}(gh)$. Because $\rho$ is a projective representation, we know these two matrices lie on the same line, hence are constant multiples of each other, but our arbitrary choice of matrix from this line may make this constant something other than one.

The collection of these nonzero constants yields a function $\phi$ from $G \times G$ to $\mathbb{C}^\times$, known as a 2-cochain. The 2-cocycle condition is given by $\frac{\phi(h,k)}{\phi(gh,k)} \frac{\phi(g,hk)}{\phi(g,h)} = 1$ for all $g,h,k \in G$, and you can check that it is satisfied here by examining lifts $\hat{\rho}$ of triple products.

If you have a 2-cocycle $\phi$ on $G$ with coefficients in $\mathbb{C}^\times$, you can make a central extension in the following way: Take the set $G \times \mathbb{C}^\times$ of pairs $(g,x)$, and twist the multiplication rule with $\phi$: $$(g,x)(h,y) = (gh, xy \cdot \phi(g,h))$$ Checking that this yields an actual group (with associative multiplcation) amounts to checking that the function $\phi$ satisfies the cocycle condition.

The second part is checking that any choice of matrices on the respective lines yields the same central extension. This arises from a notion of equivalence of cocycles known as cohomology. The equivalence classes of these cocycles form a group called the degree 2 cohomology group of $G$ with coefficients in $\mathbb{C}^\times$, and elements of this group classify central extensions.

  • $\begingroup$ That's exactly what I need to know. Thanks a lot Scott ;) $\endgroup$ Oct 25, 2011 at 20:41

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