Why is Lie algebra contraction well defined?

Question

Embarrassingly I fail to see why the following definition works (see e.g. 1)

Consider the continuous function $U:(0,1] \to GL(V)$. If the limit $\lim_{\epsilon \to 0} [x,y]_\epsilon=\lim_{\epsilon \to 0} U_\epsilon^{-1} [U_\epsilon x,U_\epsilon y]_\epsilon=[x,y]_0$ exists for all $x,y \in V$ then $[-,-]$ is a well-defined Lie bracket.

It is obvious to me why it works for any $\epsilon >0$, but as soon as $U$ is singular I do not see why all the properties of a Lie algebra are still satisfied.

E.g., Why has to be that $[x,y]_0 - [y,x]_0 =0$?

@MarianoSuárez-Álvarez to my best understanding at $\epsilon = 0$ it might be — ungerade, Aug 22 '17 at 02:55
@MarianoSuárez-Álvarez think it might (or sometimes not). It might not be a map into $GL(V)$. At least this seems to be the standard definition (see the paper in the answer). Or do i misunderstand it? — ungerade, Aug 22 '17 at 04:13
The map $U(\epsilon)$ is not defined for $\epsilon=0$, but the limit is well-defined if it exists, and "looks" like the case $\epsilon=0$. However, the limit simply may not exist. Suppose an entry in $U$ is $1/t$ and we take $t\to 0$. — Dietrich Burde, Aug 22 '17 at 08:16

Dietrich Burde · Accepted Answer · 2017-08-21T19:36:05.017

1

A Lie algebra contraction is a special case of a Lie algebra degeneration. For a Lie algebra law $\lambda$, the orbit under the above $GL(V)$-action is denoted by $O(\lambda)$. It is easy to see that a contraction of $\lambda$ lies in the orbit closure of $O(\lambda)$. By a result of Borel, the orbits are constructible sets, and hence the orbit closure with respect to the Euclidean topology is the same as the closure with respect to the Zariski topology. In other words, all invariant polynomial conditions which hold in $O(\lambda)$ also hold in the obit closure. In particular, the Jacobi identity and skew-symmetry are such polynomial conditions. Hence the law $\lambda_0\in\overline{O(\lambda)}$, given by $\lambda_0(x,y)=[x,y]_0$ is again a Lie algebra law.

For details and references see this paper.

edited Aug 21 '17 at 19:36

answered Aug 21 '17 at 18:34

Dietrich Burde

130,978

I don't understand your answer. Doesn't this follow simply from the closeness of the conditions? – Mariano Suárez-Álvarez Aug 22 '17 at 03:23
Yes, it does simply follow. The "closeness" argument here is just more general to include also degenerations, and is perhaps instructive to see. – Dietrich Burde Aug 22 '17 at 08:14

Why is Lie algebra contraction well defined?

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