why study Lie algebras? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T23:56:49Z http://mathoverflow.net/feeds/question/58696 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58696/why-study-lie-algebras why study Lie algebras? Olivier Bégassat 2011-03-17T00:24:39Z 2011-05-01T03:49:26Z <p>I don't mean to be rude asking this question, I know that the theory of Lie groups and Lie algebras is a very deep one, very aesthetic and that has broad applications in various areas of mathematics and physics. I visited a course on Lie groups, and an elementary one on Lie algebras. But I don't fully understand how those theories are being applied. I actually don't even understand the importance of Lie groups in differential geometry.</p> <p>I know, among others, of the following facts:</p> <p>$1)$ If $G$ and $H$ are two Lie groups, with $G$ simply connected, and $\mathfrak{g,h}$ are their respective Lie algebras, then there is a one to one correspondance between Lie algebra homomorphisms $\mathfrak{g}\rightarrow\mathfrak{h}$ and group homomorphisms $G\rightarrow H$.</p> <p>$2)$ The same remains true if we replace $H$ with any manifold $M$: any Lie algebra homomorphism from $\mathfrak{g}$ to the Lie algebra $\Gamma(TM)$ of smooth vector fields on $M$ gives rise to a local action of $G$ on $M$.</p> <p>$3)$ Under some conditions like (I think) compactness, the cohomology of $\mathfrak{g}$ is isomorphic to the real cohomology of the group $G$. I know that calculating the cohomology of $\mathfrak{g}$ is tractable in some cases.</p> <p>$4)$ There is a whole lot to be said of the representation theory of Lie algebras</p> <p>$5)$ Compact connected centerless Lie groups $\leftrightarrow$ complex semisimple Lie algebras</p> <p>How do people use Lie groups and Lie algebras? What questions do they ask for which Lie groups or algebras will be of any help? And if a geometer reads this, how (if at all) do you use Lie theory? How is the representation theory of Lie algebras useful in differential geometry?</p> <p>Thank you for your time</p> http://mathoverflow.net/questions/58696/why-study-lie-algebras/58700#58700 Answer by Deane Yang for why study Lie algebras? Deane Yang 2011-03-17T00:51:14Z 2011-04-30T19:08:32Z <p>Here is a brief answer: Lie groups provide a way to express the concept of a continuous family of symmetries for geometric objects. Most, if not all, of differential geometry centers around this. By differentiating the Lie group action, you get a Lie algebra action, which is a linearization of the group action. As a linear object, a Lie algebra is often a lot easier to work with than working directly with the corresponding Lie group.</p> <p>Whenever you do different kinds of differential geometry (Riemannian, Kahler, symplectic, etc.), there is always a Lie group and algebra lurking around either explicitly or implicitly.</p> <p>It is possible to learn each particular specific geometry and work with the specific Lie group and algebra without learning anything about the general theory. However, it can be extremely useful to know the general theory and find common techniques that apply to different types of geometric structures.</p> <p>Moreover, the general theory of Lie groups and algebras leads to a rich assortment of important explicit examples of geometric objects.</p> <p>I consider Lie groups and algebras to be near or at the center of the mathematical universe and among the most important and useful mathematical objects I know. As far as I can tell, they play central roles in most other fields of mathematics and not just differential geometry.</p> <p>ADDED: I have to say that I understand why this question needed to be asked. I don't think we introduce Lie groups and algebras properly to our students. They are missing from most if not all of the basic courses. Except for the orthogonal and possibly the unitary group, they are not mentioned much in differential geometry courses. They are too often introduced to students in a separate Lie group and algebra course, where everything is discussed too abstractly and too isolated from other subjects for my taste.</p> http://mathoverflow.net/questions/58696/why-study-lie-algebras/58703#58703 Answer by Scott Carter for why study Lie algebras? Scott Carter 2011-03-17T01:14:36Z 2011-03-17T01:14:36Z <p>I like Deane's answer, and I doubt that I can improve upon it, but here is an attempt. One understanding of fundamental particles is that they are representations of classical Lie groups. I think that is reason enough to study them. But more down to earth, the circle is one of the easiest examples of a Lie group to study. Its Lie algebra is the real line. The exponential map is, well, the exponential map $e^{i \theta}.$ Circles and lines are important. A next most simple example is the 3-sphere ($SU(2)$) with its Lie algebra 3-space and the Lie bracket giving $i,j,k$. These are really cool examples. The general theory might also be really cool.</p> http://mathoverflow.net/questions/58696/why-study-lie-algebras/58720#58720 Answer by Amritanshu Prasad for why study Lie algebras? Amritanshu Prasad 2011-03-17T04:14:44Z 2011-03-17T04:14:44Z <p>Lie's motivation for studying Lie groups and Lie algebras was the solution of differential equations. Lie algebras arise as the infinitesimal symmetries of differential equations, and in analogy with Galois' work on polynomial equations, understanding such symmetries can help understand the solutions of the equations.</p> <p>I found a nice discussion of some of these ideas in <em>Applications of Lie groups to differential equations</em> by Peter J. Olver, in Springer-Verlag's GTM series.</p> http://mathoverflow.net/questions/58696/why-study-lie-algebras/58733#58733 Answer by Charles Matthews for why study Lie algebras? Charles Matthews 2011-03-17T09:14:45Z 2011-03-17T11:55:53Z <p>As has been said, Lie groups are our best theory encoding continuous symmetry. Lie algebra theory, which is the infinitesimal counterpart, is a theory good enough that numerous problems can be solved by look-up, rather than arguing from first principles. You can look at the history, particularly with Cartan and Weyl; you can look at the examples coming from "commutation relations" people want to study; you can look at representation theory or root systems or the theory of universal enveloping algebras; you can look at string theory or the Langlands philosophy. It has been found very natural to look at the Lie algebra as a linearised object behind the Lie group, and something easier to study.</p> http://mathoverflow.net/questions/58696/why-study-lie-algebras/58734#58734 Answer by robot for why study Lie algebras? robot 2011-03-17T09:26:20Z 2011-03-18T00:27:39Z <p>First of all - your point 3) can be extended to a (sub)class of homogeneous spaces.</p> <p>A very nice example of a use of representation theory is the Hodge theory for Kaehler manifolds as is done e.g. in Wells's book Differential analysis on complex manifolds. On a complex manifold you have a very natural notion of $(p,q)$-forms and of $\partial$ and $\overline{\partial}$ operators. One can view this as a decomposition of exterior forms and deRham differential under a subgroup which preserves the geometric structure. But the story doesn't end here - the crucial notion of primitive cohomology is really best thought of in terms of representation theory of $\mathfrak{sl}(2,\mathbb{C})$ whose action on exterior forms commutes with the action of the structure group. </p> <p>In this example representation theory helps to organize things and calculations and there are many similar ones in spirit. E.g. orthonormal basis of harmonic functions on the sphere consisting of spherical harmonics is also an exercise in representation theory - the advantage of such a basis being the symmetry properties of its functions.</p> <p>ADDED</p> <p>Let me also try to exapand Deane Yang's answer and explain the importance of Lie groups in differential geometry. Bernhard Riemann solved the equivalence problem (i.e. the question whether a sphere is locally isometric to plane) by developing Riemann geometry and introducing the crucial invariant - the Riemann curvature. Elie Cartan developed a general method for solving such equivalence problems (see <a href="http://en.wikipedia.org/wiki/Cartan%2527s_equivalence_method" rel="nofollow">Cartan's equivalence method</a> or <a href="http://en.wikipedia.org/wiki/Method_of_moving_frames" rel="nofollow">Method of moving frames</a> on wikipedia). The notion of Lie group is already explicit there as it represents the symmetries of the geometrical structure one is interested in. This approach was later developed into what is now called <a href="http://en.wikipedia.org/wiki/Cartan_connection" rel="nofollow">Cartan geometry</a>.Informally, these geometries are curved versions of Klein geometries. The story can be told like this: </p> <p>1) classical synthetical geometry (Euclidean, projective, Lie sphere geometry, etc.)</p> <p>2a) Riemann's generalization of Euclidean geometry, introduction of manifolds </p> <p>2b) Klein's Erlangen program which postulates that every kind of geometry is determined by a homogeneous space $G/H$</p> <p>3) Cartan's generalization of these homogenous spaces in terms of $H$-principal bundles which subsumes the previous two generalizations (For details see book by <a href="http://books.google.com/books?id=Ytqs4xU5QKAC&amp;lpg=PA178&amp;dq=cartan%2520geometry%2520sharpe&amp;pg=PP1#v=onepage&amp;q=cartan%2520geometry%2520sharpe&amp;f=false" rel="nofollow">Sharpe</a>.)</p> <p>Given a geometrical structure, it is often hard theorem that the category of manifolds with this structure is isomorphic to (a certain subcategory of) the category of appropriate Cartan geometries. Nevertheless, Cartan's approach gives you very general and conceptual view on geometries like Riemannian, conformal, projective, Kaehler, quaternionic Kaehler, hyperKaehler, contact-projective, CR, ...</p> <p>Lie algebras and representation theory also appear, because the tangent space to $G/H$ can be identified with the homogeneous vector bundle associated to the $G$-representaion $\mathfrak{g}/\mathfrak{h}$ (this is one of the linearizations people keep talking about). One can regard the curvature tensor as an element of the tensor product of these and decomposition into irreducible subrepresentations then gives generalizations of Weyl and Ricci curvatures from Riemannian geometry. The Dirac operator of mathematical physics can be thought of as a deRham differential composed with a projection and an intertwining map between certain representations. In fact even such fancy gadgets as Lie algebra cohomology play their role (the keyword being "harmonic curvature").</p> <p>In the end, you see that in order to understand the appearance of Lie groups in geometry, one has to read Klein's program. The rest is just ingenious technology to allow for nonflat things. ;-)</p> http://mathoverflow.net/questions/58696/why-study-lie-algebras/58779#58779 Answer by Faisal for why study Lie algebras? Faisal 2011-03-17T19:45:38Z 2011-03-17T19:57:13Z <p>Although the title is about Lie algebras, the question body mentions Lie groups, and my answer will deal more with these. As mentioned in other answers, Lie groups show up frequently in geometry as groups of symmetries of geometric objects. For example, given a manifold $M$ we can sometimes find a Lie group $G$ that acts on $M$ in some interesting fashion, and it is then not unreasonable to hope that this action might yield information about both $G$ and $M$. </p> <p>Let's look at something a bit more specific. Suppose we have a compact connected Lie group $G$ acting 'in some nice fashion' on a manifold $M$. Typically what one does in this case is break up $M$ into $G$-orbits, and then study each piece individually. Each orbit will be a homogeneous space $G/H$ of $G$, where $H$ is the stabilizer of some point in the orbit. The space $G/H$ is very symmetric-looking, and one might try to exploit the symmetry to gain some structural information. What we have done -- roughly speaking -- is cast aside the manifold and are now working primarily with the group. Of course an interesting special case is when the action of $G$ on $M$ is transitive, i.e. when there is only one $G$-orbit in $M$ so that $M=G/H$ is itself a homogeneous space. There is so much to say about manifolds of the form $G/H$ that I will restrict myself only to two things.</p> <p><code>1)</code> The computation of the (real) cohomology of $G/H$ becomes a problem involving the Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$ of $G$ and $H$, which are <em>linear algebraic objects</em>! In particular, if $H$ is closed and connected in the compact and connected Lie group $G$ then the cohomology ring $H^\ast(G/H;\mathbb{R})$ is isomorphic to the relative Lie algebra cohomology ring $H^\ast(\mathfrak{g}, \mathfrak{h};\mathbb{R})$. For instance if $H$ is the trivial subgroup, we obtain the isomorphism $H^\ast(G;\mathbb{R}) \cong H^\ast(\mathfrak{g};\mathbb{R})$ mentioned in the OP; and indeed, computing $H^\ast(\mathfrak{g})$ is a much more tractable problem. Another interesting special case is when $H$ is a maximal torus in $G$, but I will not say more about this here...</p> <p><code>2)</code> Vector bundles over $G/H$ are related to the representation theory of $G$. Strictly speaking, this is only true of <em>equivariant</em> vector bundles, i.e. vector bundles $\pi \colon E \to G/H$ where $G$ acts on the total space $E$ in a way that respects its action on the base $G/H$: that is, we ask that $\pi(ge) = g\pi(e)$ for all $g \in G$ and $e\in E$ and that translation between fibers $E_x \to E_{gx}$ be linear. The fiber lying over the trivial coset in $G/H$ is then seen to carry a representation of $H$. Is there an action of $G$ lurking around? <em>Yes</em>: $G$ acts on the sheaf cohomology $H^\ast(G/H, V)$! Thus we can relate the cohomology of $H^\ast(G/H,V)$ to the representation theory of $G$.</p> <p>A very important special case is when $H$ is a maximal torus $T$ and $V$ is an equivariant (holomorphic) line bundle $L \to G/T$ (let's not fret about the "holomorphic" bit). (There is a miraculous fact that if $G$ is simply connected then <em>every</em> holomorphic line bundle over $G/T$ is automatically equivariant. In particular, this means that even if $G$ isn't simply connected, then we always get an action of the Lie algebra $\mathfrak{g}$ of $G$ on $H^\ast(G/H,L)$, even if there is no corresponding action of $G$. In other words, we can use the representation theory of $\mathfrak{g}$ to study $H^\ast(G/H,L)$.) There is a very explicit description of $H^\ast(G/T, L)$ in terms of the representation theory of $G$: it turns out that either $H^\ast$ vanishes completely, or else it is nonzero in a single degree $q_L$, in which case $H^{q_L}(G/T,L)$ is an irreducible representation of $G$. (This can be made much more precise; in particular, there is an explicit description of $q_L$ and of the resulting irreducible representation in terms of weights. The key phrase here is "Borel--Weil--Bott theorem.'')</p> <p>Here is a concrete example. If $G = \operatorname{SU}(2)$ and $T$ is its diagonal subgroup, then $G/T = \mathbb{C}P^1$, and one can use the Borel--Weil--Bott theorem to describe the cohomology groups $H^\ast(\mathbb{C}P^1, \mathcal{O}(n))$. For instance, the fact that $H^0(\mathbb{C}P^1, \mathcal{O}(n)) = \text{Sym}^n(\mathbb{C}^2)$ (for $n \geq 0$) comes from the fact that $\text{Sym}^n(\mathbb{C}^2)$ is the irreducible representation of $\operatorname{SU}(2)$ of highest weight $n$.</p> <hr> <p>There is another obvious reason why Lie groups are important in geometry: they are themselves geometric objects (namely, manifolds)! So you cannot expect to say something about general manifolds that cannot be said about them. Since Lie groups are a relatively well-behaved class of manifolds, one can use them as a test case of or a launch pad to more general results. The same can be said about homogeneous spaces $G/H$. For example, general results like the Atiyah--Bott fixed point foruma and the Atiyah--Singer index formulas when applied to $G/T$ (where $G$ is a compact and connected Lie group and $T$ is a maximal torus) are closely related to the Weyl character formula for $G$.</p> http://mathoverflow.net/questions/58696/why-study-lie-algebras/58784#58784 Answer by mathyperson for why study Lie algebras? mathyperson 2011-03-17T21:39:25Z 2011-03-17T21:39:25Z <p>In addition to A. Prasad's excellent recommendation (Olver's book), I would suggest you take a look at <a href="http://ocw.mit.edu/courses/mathematics/18-755-introduction-to-lie-groups-fall-2004/" rel="nofollow">Helgason's notes</a></p> <p>In particular, it is a good idea to check out <a href="http://ocw.mit.edu/courses/mathematics/18-755-introduction-to-lie-groups-fall-2004/readings/" rel="nofollow">the bottom of the page here.</a> The last three papers in the additional readings section gives a non-technical account of the the origins of Lie groups. </p> http://mathoverflow.net/questions/58696/why-study-lie-algebras/58808#58808 Answer by Emerton for why study Lie algebras? Emerton 2011-03-18T03:39:57Z 2011-03-18T03:39:57Z <p>Here is a very fundamental way to create interesting Riemannian manifolds: Let $G$ be a semi-simple Lie group, let $K$ be its maximal compact subgroup, let $\Gamma$ be a discrete subgroup of $G$, and form $G / K.$ This quotient is called the symmetric space attached to $G$.</p> <p>The Riemanian structure comes from an invariant metric on $G$, and so $G$ acts as isometries on $G/K$ by left translation.</p> <p>If you consider the case $G = SL_2(\mathbb R)$, you get $SL_2(\mathbb R)/SO(2)$, which is naturally identified with the complex upper half-plane (on which $SL_2(\mathbb R)$ acts via Mobius transformations; note that the point $i$ is stablized precisely by $SO(2)$), which is also the hyperbolic plane. Other groups give higher dimensional hyperbolic spaces (e.g. $SL_2(\mathbb C)$ gives hyperbolic $3$-space), the Siegel upper half-spaces (from symplectic groups), complex balls, and many other well-known spaces.</p> <p>If you now take a discrete subgroup $\Gamma$ of $G$, you can form the double quotient $\Gamma \backslash G /K$. These are some of the most celebrated Riemannian manifolds in mathematics. In the case of $SL_2(\mathbb R)$, we know via uniformization that all genus $\geq 2$ Riemann surfaces can be described in this way. In the case of $SL_2(\mathbb C)$ we get hyperbolic $3$-manifolds, from symplectic groups we get moduli spaces of abelian varieties, ... .</p> <p>Now (as the preceding discussion hopefully makes clear), lots of these spaces are known by other names that don't involve Lie theory, and can be studied in a non-Lie-theoretic way. But the Lie-theoretic perspective provides a unifying, and frequently clarifying, point of view. For example, cohomological or function-theoretic invariants of these spaces can often be described and computed via Lie theoretic tools (e.g. via Lie algebra cohomology of certain unitary representations of the group $G$).</p> <p>As a concluding remark, let me note that a general principle is that when certain symmetries are implict in a given context (e.g. $SL_2(R)$ being the group of hyperbolic isometries of the upper half-plane), it is good to explicitly bring them to the fore and take them into account. In geometry, the symmetry groups that appear (of a space, or perhaps of its universal cover) are very often Lie groups. And so a little knowledge of Lie theory can turn into a powerful tool for investigating a given geometric situation.</p> <p>P.S. I should also note that the study of spaces $\Gamma \backslash G/K$ for certain $\Gamma$ (so-called congruence subgroups) is one of the basic topics of the Langlands program, and the function theory and cohomology of these spaces (especially their representation-theoretic structure) is conjectured to govern a vast amount of number theory. Trying to understand and work on these conjectures was my own motivation for learning Lie theory.</p> http://mathoverflow.net/questions/58696/why-study-lie-algebras/63546#63546 Answer by Dinesh for why study Lie algebras? Dinesh 2011-04-30T17:44:44Z 2011-04-30T17:44:44Z <p>@Amritanshu Is this related to differential Galois theory? I think Differential Galois Theory answers why some linear differential equations are solvable in known functions..does lie groups used to solve differential equations have anything to say about this?</p> http://mathoverflow.net/questions/58696/why-study-lie-algebras/63577#63577 Answer by Igor Belegradek for why study Lie algebras? Igor Belegradek 2011-05-01T03:49:26Z 2011-05-01T03:49:26Z <p>Large subfields of modern differential geometry hardly ever use Lie group theory, e.g. they are never mentioned (as far as I can see) in Schoen-Yau's "Lectures on Differential Geometry", and their role in comparison geometry is quite modest. Major uses of Lie groups in Riemannian geometry are: </p> <ol> <li><p><a href="http://en.wikipedia.org/wiki/Holonomy" rel="nofollow">Holonomy groups.</a> </p></li> <li><p>Principal bundles and <a href="http://en.wikipedia.org/wiki/Chern%E2%80%93Weil_homomorphism" rel="nofollow">Chern-Weil theory.</a></p></li> <li><p>Homogeneous and symmetric spaces, as a source of fundamental examples of Riemannian manifolds.</p></li> <li><p>Collapsing theory with two sided curvature bounds (where local models are nilpotent Lie groups).</p></li> </ol> <p>Kobayashi-Nomizu's two volume "Foundations of Differential Geometry" discusses 1,2,3 extensively.</p>