Let $\frak{g}$ be the (real) Lie algebra:

$${\frak g}=\{A =M_n(\mathbb{R}) \mid \text{tr}(A)=0\}$$

with Lie bracket $[x,y]=xy-yx$ for $x,y\in \frak g$

Than how do I construct the corresponding Lie group by using the exponential map?

  • $\begingroup$ Can you be more explicit? So the set $\{\exp(A) \mid A \in \frak sl_n \}$ will form a Lie group with matrix multiplication? $\endgroup$ – SamC Sep 23 '15 at 9:44

Let's collect a few facts:

(i) For any matrix $A$, $e^A$ is well defined.

(ii) We have $\det e^A = e^{\mathrm{Tr} A}$. Hence exponential map takes Lie algebra of all matrices $\mathfrak{gl}(n)$ into group of nonsingular matrices $\mathrm{GL}(n)$. Furthermore $\mathrm{Tr}A=0$ if and only if $\det e^A =1$.

From that you can see that your Lie algebra, known as $\mathfrak{sl}(n)$ is Lie algebra of $\mathrm{SL}(n)$ - group of matrices of unit determinant.


As Dietrich Burde correctly pointed out, group you are looking for is not uniquely determined. Given two groups with isomorphic Lie algebras you can only infer that they are isomorphic in some neighbourhood of identity. Well known examples of this phenomenon are: $SU(2)$, $SO(3)$, $O(3)$. They all have the same Lie algebra. However $O(3)$ is distinct from $SO(3)$ as it contains two disjoint connected components (both diffeomorphic to $SO(3)$). $SU(2)$ on the other hand is connected and even simply-connected (which $SO(3)$ is not). In fact it is double cover of $SO(3)$. Even simpler example is circle group $SO(2) \cong U(1) \cong S^1$ which is locally isomorphic to $\mathbb R$ with addition. In fact $\mathbb R$ is simply connected cover of circle group. So without some extra conditions you can't quite pin down what group is that.

  • $\begingroup$ However, $\exp: \mathfrak{sl}_n(\mathbb{R})\mapsto SL_n(\mathbb{R})$ is not surjective. $\endgroup$ – Dietrich Burde Sep 23 '15 at 12:45
  • $\begingroup$ Well if we come to that then sure, group that SamC is interested in is not uniquely determined. We would need some extra information to uniquely choose one. If we don't have it, $\mathrm{SL}(n)$ is as good representative as any. Note also that exponential map will give us at least some open neighbourhood of identity in $\mathrm{SL}(n)$ and that already generates whole group. $\endgroup$ – Blazej Sep 23 '15 at 13:18
  • $\begingroup$ I would like to see some development on the title of the OP: is there a general way to compute the Lie group from the algebra? For instance, in Hall's Lie Groups, Lie Algebras, and Representations, see Definition 3.18: the Lie algebra is the set of matrices $X$ such that $e^{tX}$ is in G for real $t$. So it is not so simple as exponentiate matrices in the algebra. At least can you retrieve the whole connected component of the identity or are you restricted to a neighbourhood? $\endgroup$ – Miguel Sep 23 '15 at 14:03
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    $\begingroup$ The question has been discussed also here. $\endgroup$ – Dietrich Burde Sep 23 '15 at 18:13
  • $\begingroup$ In general you won't get whole connected component by exponentiation. However every neighbourhood of identity generates the group (I assume connectedness now), i.e. given neighbourhood of identity $e\in U\in Top(G)$ every element of group can be written as finite product of elements of $U$. In particular every element of group can be written down as finite product of exponentials $e^{X_1}e^{X_2}...e^{X_n}$. $\endgroup$ – Blazej Sep 23 '15 at 22:57

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