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Background

Let $(M,g)$ be a finite-dimensional riemannian (or more generally pseudoriemannian) manifold. Suppose that I know that a certain Lie group $G$ acts transitively and isometrically on $M$ and after a little bit more work I exhibit $M$ as a homogeneous space $G/H$. Let $$ \mathfrak{g} = \mathfrak{h} \oplus \mathfrak{m}~, $$ where $\mathfrak{g}$ and $\mathfrak{h}$ are the Lie algebras of $G$ and $H$ respectively, and suppose that this split is reductive. (This is a nonempty condition in the pseudoriemannian case.)

The group $G$ is a subgroup of the isometry group of $M$, but it need not be the full isometry group. For example, we could have $M$ be a compact Lie group $G$ with the natural bi-invariant metric coming from (minus) the Killing form on the Lie algebra. Clearly $G$ acts transitively on $G$ via left multiplication, but the full isometry group is $G \times G$ (modulo the centre, if we insist that the isometry group acts effectively).

Question

Is there a way to determine the isometry Lie algebra of a homogeneous pseudoriemannian manifold $G/H$, preferably by algebraic means from the data $(\mathfrak{g},\mathfrak{h})$ and the $\mathfrak{h}$-invariant inner product on $\mathfrak{m}$?


This is not idle curiousity, by the way. In some work I'm doing, I have encountered an explicit 7-dimensional homogeneous lorentzian manifold which I can describe as $G/H$, but it is important that I know whether the isometry Lie algebra is indeed $\mathfrak{g}$ or something larger. I fear, though, that writing down the explicit example I am faced with might be deemed "too localized" (and rightly so), and I'm hoping this more general question is acceptable.

Update

I have now ran the algorithm in Robert's answer for my 7-dimensional homogeneous space in my work and out popped one additional Killing vector field! (Not the result I wished for, but explains something I did not understand.)

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I know only about the Riemannian case. Of course the case of symmetric spaces is well known, going back to Cartan, and then there is more recent work of Onishchik. But the nicest result I know of is even more recent and due to S. Regianni arXiv:0906.1802v3 [math.DG]. He describes the isometry group of compact Riemannian naturally reductive spaces. I am not sure it is useful to you. –  Claudio Gorodski Sep 19 '11 at 20:35

1 Answer 1

up vote 20 down vote accepted

Here is an algorithm to compute the Lie algebra of the group of isometries of a homogeneous space $G/H$ endowed with a $G$-invariant (pseudo-)Riemannian metric $g$. It is phrased in terms of essentially algebraic computations using the left-invariant forms on $G$, but it could be reduced completely to computations with the Lie algebra (and the matrix $Q$ that defines the metric) if that is what one wanted to do.

Let $\frak{g}$ and $\frak{h}\subset\frak{g}$ denote the Lie algebras of $G$ and $H\subset G$ respectively. Set $s = \dim\frak{h}$ and let $n>0$ be the dimension of $\frak{m} = \frak{g}/\frak{h}$.

Let $\omega = (\omega^i)$ (where $1\le i\le n$) be a basis for the left-invariant $1$-forms on $G$ such that $\omega = 0$ defines the foliation of $G$ by the left cosets of $H$. Then there will exist a unique non-degenerate, symmetric $n$-by-$n$ matrix $Q$ such that the quadratic form $Q_{ij}\omega^i\circ\omega^j$ is the $\pi$-pullback to $G$ of $g$ under the natural coset mapping $\pi:G\to G/H$.

Since $g$ is invariant under the action of $G$, there exists a unique $n$-by-$n$ matrix $\theta = (\theta^i_j)$, whose entries are left-invariant $1$-forms on $G$, such that $$ d\omega = {}-\theta\wedge\omega \qquad\text{and}\qquad Q\theta + {}^t(Q\theta) = 0. $$ (This is just the Fundamental Lemma of (pseudo-)Riemannian Geometry in this context.) Note that applying $d$ to both sides of this equation gives $0 = \Theta\wedge\omega$, where, of course, $\Theta = d\theta + \theta\wedge\theta$, is the curvature of the connection $\theta$ and hence can be written in the form $\Theta = R(\omega\wedge\omega)$, where the coefficients in $R$ are constants, since $R$ is left-invariant as a function on $G$.

Suppose now that a vector field $Y$ on $G$ be $\pi$-related to a $g$-Killing vector field $Z$ on $G/H$, and let $\omega(Y) = a$. Then the $g$-Killing equation for $Z$ implies that $$ da = {}-\theta\ a + b\ \omega $$ where $b$ is an $n$-by-$n$ matrix of functions that satisfies $Qb + {}^t(Qb)=0$. Taking the exterior derivative of this equation gives $$ 0 = {} -\Theta\ a + (db +\theta\ b - b\ \theta)\wedge\omega. $$ Now, by counting dimensions to show that the corresponding linear algebra problem always has a unique solution, it is easy to see (and, in any given case, explicitly compute) that there exists a matrix $\rho = (\rho^i_j)$ of $1$-forms such that $\Theta\ a = \rho\wedge\omega$ and such that $Q\rho + {}^t(Q\rho) = 0$. In fact, one has $\rho^i_j = c^i_{jkl}a^k\omega^l$, where the $c^i_{jkl}$ are constants determined by the curvature form $\Theta$. (The exact formula is not important for the following argument.) Thus, the above equation can be written as $$ db = -\theta\ b + b\ \theta + \rho(a,\omega), $$ where I have written the term $\rho$ as $\rho(a,\omega)$ to emphasize that this is some constant-coefficient bilinear pairing of $a$ and $\omega$ taking values in the Lie algebra ${\frak{so}}(Q)$.

The above equations for $da$ and $db$ are then a total (linear) differential system whose solutions give the Lie algebra of $g$-Killing vector fields on $G/H$.

Now, this system is not Frobenius unless the $(G/H,g)$ is a (pseudo-)Riemannian space form, so you have to differentiate these equations. The derivative of the $da$-equation won't give anything new, so one must differentiate the $db$-equation. This yields equations of the form $$ 0 = d(db) = B, $$ where $B = (B^i_j)$ and $B^i_j = (b^i_{jklm}a^m + c^{ip}_{jklq}b^q_p)\omega^k\wedge\omega^l$ for some explicit constants $b^i_{jklm}= -b^i_{jlkm}$ and other constants $c^{ip}_{jklq} = - c^{ip}_{jlkq}$. (I can't see anything wrong with the TeX here, but it's not typesetting correctly.)

It follows that the $a^i$ and $b^i_j$ are subject to the constant-coefficient linear relations $b^i_{jklm}a^m + c^{ip}_{jklq}b^q_p = 0$ in addition to the linear relations on $b$ already known: $Qb + {}^t(Qb) = 0$. Differentiating these new linear relations and using the $da$ and $db$ formulae to express the results in terms of left-invariant forms with coeffcients that are constant linear combinations of the $a$- and $b$-components, one might get further constant coefficient linear relations among the $a$- and $b$-components. Repeat this process with the new relations (if any) until no new linear relations are found.

At this point, the linear relations between the $a$- and $b$-components will have some solution space of dimension $n{+}s{+}r$ for some $r\ge0$ (but, necessarily, $r\le n(n{-}1)/2 - s$). It will then follow that the space of $g$-Killing vector fields on $G/H$ has dimension $n{+}s{+}r$. Moreover, one can compute the Lie algebra structure on this space by using the formula $$ \omega\bigl([Y_1,Y_2]\bigr) = Y_1(a_2) - Y_2(a_1) - \theta\wedge\omega\bigl(Y_1,Y_2\bigr) $$ and the formulae for $da_1$ and $da_2$. Thus, the algebra structure of the $g$-Killing fields will follow directly by algebraic operations from the structure of the algebras $\frak{g}$ and $\frak{h}$ and $Q$.

In any given instance, this algorithm can be implemented on a computer without difficulty.

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Dear Robert, Many thanks for this beautiful answer! This is exactly what I needed. Thanks, José –  José Figueroa-O'Farrill Sep 19 '11 at 21:25
    
Dear Robert -- 1. I'd like to second Jose's comment. 2. I've also tried to fix the formula that did not show correctly by inserting back ticks around it. It seems to have worked but the formula became slightly smaller. –  algori Sep 19 '11 at 21:33
    
You're welcome. Looking it over, I realize that I forgot to say that I was assuming that $G$ acts almost effectively on $G/H$. If that's not the case, then you have to let $s$ be the dimension of $H/N$, where $N\subset H$ is the ineffective subgroup of the $G$-action on $G/H$. –  Robert Bryant Sep 19 '11 at 23:00
    
Dear Robert, Thanks again. I found an additional Killing vector in my example and this explains something that I didn't understand (and why I asked the question in the first place). I wish I could upvote your answer again. –  José Figueroa-O'Farrill Oct 2 '11 at 21:21
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I'm glad you found it useful. It's always a pleasure to be be able to help (which is what makes MO work, when you think about it). –  Robert Bryant Oct 2 '11 at 22:15

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