# Nilpotent Lie algebras and unipotent Lie groups

$\mathbf{n}$ is nilpotent Lie algebra with $N$ being the corresponding algebraic Lie group. Now one neat feature of this setting is that you can take the exponential map to be identity. In other words you can define a group structure on $\mathbf{n}$ using the Campbell-Hausdorff formula. I have the following questions (they might be very easy; I just don't know):

1. If $\pi: \mathbf{n} \rightarrow \text{End}(V)$ is a representation of the nilpotent Lie algebra then does $x \in N$ acts as $\exp (\pi (x))$ on $V$ ?
2. If $\mathbf{m}$ is a subalgebra of $\mathbf{n}$ then you get a corresponding group $M \subset N$. Do we have: $N/M = \mathbf{n}/\mathbf{m}$ as sets?

EDIT: Victor's comments are to the point so in order to clear up the confusion I added that $N$ is algebraic.

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This is a very confused post. (1) In Lie theory and differential geometry, the term "exponential map" has a precise meaning and, no, the exponential map is not an identity in your situation. For example, let $N$ be the circle group, which is even abelian, then the exponential map is not injective. (2) The Baker-Campbell-Hausdorff formula is typically used to describe the group structure on $N$ in exponential coordinates, not to define "group structure" on $n$ by pull-back by some map $n\to N.$ – Victor Protsak Sep 11 '10 at 3:32
(3) For any Lie group $G$, whether nilpotent or not, if $g\in G$ is in the image of the exponential map, $g=\exp(X),$ then it follows by the definition of the Lie correspondence between representations of $G$ and $Lie(G)$ that $\pi(g)=\exp(\pi(X)).$ (4) The circle group example shows that $N/M$ need not be contractible (take $N=S^1$ and $M=\{1\}$). I suggest working through a good introductory book on Lie theory that discusses the correspondence between Lie groups and Lie algebras. – Victor Protsak Sep 11 '10 at 3:37
You probably want to add the condition that N is simply connected. – Richard Borcherds Sep 11 '10 at 4:39
@Victor sorry. I added the condition, $N$ being algebraic. @Richard I thought unipotent groups were simply connected. – Najdorf Sep 11 '10 at 5:00

I will suppose that $\mathfrak n$, etc is finite-dimensional.

If by "the corresponding Lie group" you mean "the corresponding connected simply-connected Lie group", then what you say is correct (in spite of Victor Prostak's comment). Let $\{\mathfrak n,[,]\}$ be a Lie algebra over a field $k$ of characteristic $0$. Then the BCH formula truncates, and in particular converges. In general, BCH defines a "partial group" structure in its domain of convergence (as for "typical uses", in finite dimensions over $\mathbb R$, BCH has positive radious of convergence, and by gluing together such patches one can prove the Lie III theorem), and so for nilpotent $\mathfrak n$ it defines a group structure $\{\mathfrak n, \operatorname{BCH}\}$ on the set of points in $\mathfrak n$. In fact, when $\{\mathfrak n,[,]\}$ is nilpotent, the fact that BCH truncates means that it is polynomial, and so $\{\mathfrak n, \operatorname{BCH}\}$ is affine algebraic over $k$. In particular, when $k = \mathbb R$, then BCH defines a real algebraic group $\{\mathfrak n, \operatorname{BCH}\}$, and in particular a Lie group, and it is straightforward to check that $\operatorname{Lie}(\{\mathfrak n, \operatorname{BCH}\}) = \{\mathfrak n,[,]\}$. Since as a topological space $\mathfrak n$ is connected and simply connected, $\{\mathfrak n, \operatorname{BCH}\}$ is the connected simply-connected group corresponding to $\{\mathfrak n,[,]\}$. This I think you already know, but I thought it best to spell it out for other readers.

(1.) Yes, if I'm understanding correctly. Let $\pi : \{\mathfrak n,[,]\} \to \operatorname{End}(V)$ be a Lie algebra homomorphism, and $\dim V < \infty$. Since $\{\mathfrak n, \operatorname{BCH}\}$ is connected and simply connected, $\pi$ determines a group homomorphism $\Pi: \{\mathfrak n, \operatorname{BCH}\} \to \operatorname{GL}(V)$. Given $x\in \mathfrak n$, yes, $\Pi(x) = \exp(\pi (x))$.

There's nothing special about unipotent groups here. Let $\mathfrak g$ be a (finite-dimensional) Lie algebra (over $\mathbb R$) and $G$ the corresponding connected simply-connected Lie group, and let $\exp : \mathfrak g \to G$ be the exponential map (in your example, $\exp: \{\mathfrak n,[,]\} \to \{\mathfrak n, \operatorname{BCH}\}$ is an isomorphism of spaces, so we call it the identity). Let $H$ be any other Lie group and $\mathfrak h = \operatorname{Lie}(H)$, and let $\exp: \mathfrak h \to H$ be the exponential map. Then there is a canonical bijection between Lie algebra homomorphism $\mathfrak g \to \mathfrak h$ and Lie group homomorphisms $G \to H$, and the bijection intertwines the two exponential maps ($\Phi\circ \exp = \exp \circ \phi$). One proof is uses the following description of $\exp$: we can identify $\mathfrak g \hookrightarrow \Gamma({\rm T}G)$ as left-invariant vector fields, so that $x\in \mathfrak g$ determines an ODE on $G$, and it turns out that this ODE has solutions for all times, and $\exp(x)$ is the image of $e\in G$ under the flow-by-time-$1$ map.

(2.) No. I think what you are asking is the following. Let $\{\mathfrak n,[,]\}$ be a (finite dimensional) nilpotent Lie algebra (over $\mathbb R$) and $\{\mathfrak m,[,]\}$ a subalgebra. Then we have connected simply-connected groups $\{\mathfrak n,\operatorname{BCH}\}$ and $\{\mathfrak m,\operatorname{BCH}\}$, and the latter is canonically a subgroup of the former. Then I think what you are asking is whether the cosets of $\{\mathfrak m,\operatorname{BCH}\}$ in $\{\mathfrak n,\operatorname{BCH}\}$ are the same as the cosets of the vector space $\{\mathfrak m,+\}$ in the vector space $\{\mathfrak n,+\}$.

Consider the three-dimensional Heisenberg algebra, which is generated by two elements $x,y$, with the condition that $[x,y]$ is central (then $x,y,[x,y]$ form a basis). This has the matrix representation: $$x = \left( \begin{array}{ccc} 0 & 1 & 0 \\ & 0 & 0 \\ & & 0 \end{array} \right), \quad y = \left( \begin{array}{ccc} 0 & 0 & 0 \\ & 0 & 1 \\ & & 0 \end{array} \right), \quad [x,y] = \left( \begin{array}{ccc} 0 & 0 & 1 \\ & 0 & 0 \\ & & 0 \end{array} \right)$$ Then $\operatorname{BCH}(ax,y) = ax + y + \frac a 2 [x,y]$. In particular, letting $\mathfrak n$ be the Heisenberg algebra and $\mathfrak m$ the one-dimensional subalgebra spanned by $x$, we see that the (right, say) cosets of $\{\mathfrak m,\operatorname{BCH}\}$ do not agree with the cosets of $\{\mathfrak m,+\}$: in particular, the cosets through $y$ are different (one is a line parallel to the span of $x$, the other is a line parallel to the span of $x + \frac12 [x,y]$).

Note that as Emerton points out, if $\{\mathfrak m,[,]\}$ is a Lie ideal in $\{\mathfrak n,[,]\}$, then the answer is yes. Let $y \in \mathfrak n$. Then the coset of $\{\mathfrak m,\operatorname{BCH}\}$ through $y$ consists of elements of the form $x + y + l$, where $x \in \mathfrak m$ and $l$ is a Lie polynomial with at least one $x$, and hence (since $\mathfrak m$ is an ideal), it is in $\mathfrak m$. I.e. the coset is precisely $y + \mathfrak m$.

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By "what you say is correct", I mean that I started writing this before you made any edits; if you are working over $\mathbb R$, you can do what you seem to want to without worrying about algebraic groups. Note that the circle $S^1$ is a real algebraic group (for example, $\operatorname{SO}(2,\mathbb R)$), but the exponential map from $\operatorname{Lie}(S^1) \to S^1$ is not an algebraic map. So exponential maps always exist in smooth land, but not always in algebraic land. But the BCH series shows that any nilpotent algebra does integrate to an algebraic group for which $\exp$ is algebraic. – Theo Johnson-Freyd Sep 11 '10 at 5:18

Dear Najdorf,

In light of Victor Prostak's comments, it might be good to remark that the correspondence you are thinking of is between nilpotent Lie algebras and unipotent algebraic groups (over a field of characteristic zero). It is then true that you can identify $\mathfrak n$ with $N$, and use the Cambell-Hausdorff structure to define the group law on $N$.

As Victor notes, the answer to question (1) is yes: one direct way to see it is that the image of $\mathfrak n$ in $End(V)$ is again nilpotent, and then to note that the construction of $N$ from $\mathfrak n$ is functorial (since it is just a matter of using the CH formula).

As for (2), if $\mathfrak m$ is an ideal in $\mathfrak n$ (equivalently, if $M$ is normal in $N$), then the answer is yes. (The quotient $\mathfrak n/\mathfrak m$ is again a nilpotent Lie algebra, and the associated unipotent group is canonically identified with $N/M$.)

If $\mathfrak m$ is just a Lie subalgebra, then I'm less sure: without the ideal assumption, it doesn't seem that cosets under adding $\mathfrak m$ and cosets under multiplying by $M$ are the same, and so it's not clear (to me) how one would make this identification.

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Thank you I added the fact that $N$ is algebraic. – Najdorf Sep 11 '10 at 5:02