Conformal algebra in 2 dimensions admits a central extension and while the procedure itself is explained in many textbooks quite well, I haven't found a source in which it is reasoned as to why we should carry out the central extension.    In  Wikipedia article, it is stated that the symmetry group of a quantized system usually is a central extension of the classical symmetry group, and in the same way the corresponding symmetry Lie algebra of the quantum system is, in general, a central extension of the classical symmetry algebra.  Can someone please elaborate on this and explain why this is the case?
103163
There seem to be two different things one must consider when representing a symmetry group in quantum mechanics:   The universal cover: For instance, when representing the rotation group $\mathrm{SO}(3)$, it turns out that one must allow also $\mathrm{SU}(2)$ representations, since the negative sign a "$2\pi$ rotation" induces in $\mathrm{SU}(2)$ is an overall phase that doesn't change the physics. Equivalently, all representations of the Lie algebra are what we seek. ($\mathfrak{so}(3) = \mathfrak{su}(2)$, but although every representation of the algebra is one of the universal cover, not every representation of the algebra is one of $\mathrm{SO}(3)$.) Central extensions: In conformal field theory, one has classically the  of infinitesimal conformal transformations. From the universal cover treatment one is used to in most other cases, one would expect nothing changes in the quantum case, since we are already seeking only representation of an algebra. Nevertheless, in the quantization process, a  appears, which is often interpreted to arise as an "ordering constant" for the now no longer commuting fields, and we have to consider the  instead.   The question is: What is going on here? Is there a way to explain both the appearence of universal covers and central extensions in a unified way?
68471
I have been reading about multi-dimensional numbers, and found out that it's been proven that the Octonions are the composition algebra of the largest dimension. I was wondering why, despite having infinitely many different dimensions of numbers, the only composition algebras are of 1, 2, 4, and 8 dimensions. What's so special about 8?
11258
I am trying to find all 1- or 2- dimensional Lie Algebras "a" up to isomorphism. This is what I have so far:  If a is 1-dimensional, then every vector (and therefore every tangent vector field) is of the form $cX$. Then , by anti-symmetry, and bilinearity:  $$[X,cX]=c[X,X]= -c[X,X]==0$$  I think this forces a unique Lie algebra because Lie algebra isomorphisms preserve the bracket.  I also know Reals $\mathbb{R}$ are the only 1-dimensional Lie group, so its Lie algebra ($\mathbb{R}$ also) is also 1-dimensional. How can I show that every other 1-dimensional algebra is isomorphic to this one? Do I use preservation of bracket?  For 2 dimensions, I am trying to use the fact that the dimension of the Lie algebra g of a group $G$ is the same as the dimension of the ambient group/manifold $G$. I know that all surfaces (i.e., groups of dimension 2) can be classified as products of spheres and Tori, and I think the only 2-dimensional Lie group is $S^1\times S^1$, but I am not sure every Lie algebra can be realized as the Lie algebra of a Lie group ( I think this is true in the finite-dimensional case, but I am not sure).  I know there is a result out there that I cannot yet prove that all 1- and  2-dimensional Lie algebras are isomorphic to Lie subalgebras of $GL(2,\mathbb{R})$ (using matrix multiplication, of course); would someone suggest how to show this last? Thanks.
106633
Exercise 14.33 in Fulton and Harris's Representation Theory claims that  If the roots of a semisimple Lie algebra lie in a collection of mutually orthogonal subsets, one sees that the Lie algebra decomposes into a direct sum accordingly.  Although intuitively I can see why this is true, could anybody direct me to a proof of this assertion, or outline how I should go about doing it?
42445
I've been trying to find an example of a non abelian, non connected Lie group such that its Lie algebra is abelian but I can't find one, so can somenone please help me.
3917
I'd like to know a reference for a simple proof that $\{c_\mu\mid \mu\vdash n\}$ is a basis for the centre of the symmetric group algebra $\mathbb{C}\mathfrak{S}_n$, where $c_\mu$ is the sum of all elements of a given cycle type (i.e. the sum of every element in the conjugacy class corresponding to $\mu$).
94310
I have been attempting this problem for a while, it is an assignment problem so I don't want somebody to just post the answer, I'm just looking for hints.  Let $\mathfrak{u}(n,\mathbb{C})$ be the Lie algebra of strictly upper triangular $n\times n$ matrices over $\mathbb{C}$.  I am trying to show that $\mathfrak{u}(n,\mathbb{C})$ is a nilpotent Lie algebra for $n\geq 2$.  This is what I have looked at so far.  I have done some explicit calculations (I shall omit the bulk of them and just summarise below) for $n=2,3,4$ and have come up with some ideas.  Let $L=\mathfrak{u}(n, \mathbb{C})$.  If $n=2$,  $L^1 = [L, L] = &lt;0&gt;_{\mathbb{C}}$  So $L$ is nilpotent.  If $n=3$,  $L^1 = [L, L] = &lt;e_{13}&gt;_{\mathbb{C}}$  $L^2 = [L, L^1] = &lt;0&gt;_{\mathbb{C}} = \left\{ 0 \right\}$  So, $L$ is nilpotent.  If $n=3$  $L^1 = [L,L] = &lt;e_{13},e_{14},e_{24}&gt;_\mathbb{C}$  $L^2 = [L,L^1] = &lt;e_{14}&gt;_{\mathbb{C}}$  $L^3 = [L, L^2] = &lt;0&gt;_{\mathbb{C}} = \left\{ 0 \right\}$  So, $L$ is nilpotent.  So my hypothesis so far is that for an arbitrary $m\in\mathbb{N}$, $L^{m-1}=\{0\}$  My instinct is to do this by induction. My base case, $n=2$ has already been shown. Now I assume that $\mathfrak{u}(n,\mathbb{C})$ is nilpotent for $n=k$ ($k\in\mathbb{N}$) and consider the case for n=k+1.  However, I have no idea how to link these two cases together. I have read in Introduction to Lie Algebras and Representation Theory - J.E. Humphreys that $L^k$ should be spanned by basis vectors $e_{ij}$ where $j-i=k+1$ - however in my calculations for $L^1$, I obtain an $e_{14}$ which does not satisfy 4-1=2.  Am I missing something completely obvious? Or have I made a mistake somewhere? (I have checked my calculations numerous times, but cannot find any errors!)  Thanks,  Andy.
123556
I was hoping someone could give an overview as to how the Lie groups $SO(3)$ and $SU(2)$ and their representations can be applied to describe particle physics?  The application of Lie groups and their representations is an enormous field, with vast implications for physics with respect to such things as unification, but I what specifically made these groups of physical importance and why there study is useful.  I have just started studying these two groups in particular, but from a mathematical perspective, I'd very much appreciate understanding some sort of physical motivation.
122940
Why is  quaternion algebra 4D and not 3D?  Complex algebra is 2D and what is known as quaternion algebra jumps to 4D.    $ i^2 = j^2 = k^2 = ijk = -1 $  Using $1, i, j,$ and $k$ as the base (where complex uses $1$ and $i$ (or $j$ if you are an EE)) which results in a 4-axis space.  Why is there no 3D algebra?
37167
I don't get a grip of what that exactly means. What IS an abstract singlet, doublet,... under $SU(N)$ or other groups?
68976
Is there an easy way to prove that every Lie subgroup of $SU(2)$ is closed? That is, is it true that any $K=\exp_{SU(2)}(k)$ is closed for every subalgebra $k \subset \mathfrak{su}(2)$?  If there isn't, is there a reference for this fact?  I thought that this was left as an exercise on the book on compact Lie groups by Sepanski, but I cannot find the reference. Thanks.
21273
Let $G$ be a matrix Lie group. Its Lie algebra $\mathfrak{g}$ comes equipped with a symmetric (2,0) tensor known as its Killing from, denoted $K$ and defined by  $$ K(X,Y) = -\text{Tr}(\text{ad}_X \circ \text{ad}_Y)$$  where $\text{ad}_X \in \text{GL}(\mathfrak{g})$ defined by $\text{ad}_X(Y)= [X,Y]$. Show that $K$ is Ad-invariant, in the sense that  $$  K(\text{Ad}_A(X), \text{Ad}_A(Y)) = K(X,Y) \quad \forall X,Y \in \mathfrak{g}, A \in G $$.  Here $\text{Ad}_A(X) = AXA^{-1} \in \mathfrak{g}$.   I suppose that it suffices to prove it on a basis, but it isn't a big deal to let $X$ and $Y$ be arbitrary. Choosing a basis $E_j$ for $\mathfrak{g}$, I write $$ [E_j,E_k] = {c_{jk}}^l E_l $$ $$ X = a^kE_k, Y = b^kE_k $$ and  $$ AE_kA^{-1} = {d_k}^lE_l. $$ Then $$ [X,[Y,E_m]] = [a^kE_k,[b^lE_l,E_m]] = a^kb^l {c_{kn}}^j {c_{lm}}^n E_j$$ and so the trace of this operator is $a^kb^l{c_{kn}}^j{c_{lj}}^n$. The left hand side is $$  [AXA^{-1},[AYA^{-1},E_m]] = [a^k{d_k}^lE_l,[b^n {d_n}^j E_j,E_m]]= a^kb^n{d_k}^l{d_n}^j {c_{jm}}^r {c_{lr}}^k E_k$$ so the trace of this operator is $$ a^kb^n{d_n}^j{d_k}^l{c_{jk}}^r{c_{lr}}^k $$  assuming the indices are correct. Now these two have to be the same somehow. Using that $\text{Ad}_A{[X,Y]}=[\text{Ad}_A{X}, \text{Ad}_A{Y}]$ may be promising, but I haven't found anything useful with it. A hint would be appreciated.
100419
Almost all standard sources on the existence of Majorana spinors (e.g. Appendix B.1 to Polchinski's "String Theory", Vol. 2) do so in a way I consider inherently ugly:   A priori, we are dealing with an irreducible complex representation $(V,\rho)$ of the Clifford algebra of signature $(p,q)$, i.e. generalized Dirac spinors. That Majorana spinors exist means abstractly that there is a  on $V$, i.e. a conjugate-linear map $\phi : V\to V$ with $\phi^2 = \mathrm{id}_V$ that commutes at least with the $\mathfrak{so}(p,q)$ action.  Every single source I can find for Majorana spinors uses operations like the transpose, complex conjugation and Hermitian adjoint on the $\Gamma$-matrices to obtain matrices acting on the same space. This is abstractly wrong, the transpose acts on the dual, the complex conjugation on the conjugate, and the Hermitian adjoint needs an inner product we have no reason for choosing. Of course, since $V$ is finite-dimensional, one can pick a basis and define the uncanonical isomorphisms to its dual and its conjugate, but I find this inelegant, particularly since the standard derivations require us to make a particular such choice with respect to the signs the $\Gamma$-matrices have under e.g. ${}^\dagger$. Finally, the Majorana spinors are usually defined by some equation involving an unnatural and arbitrary-looking product of $\Gamma$-matrices, which varies from source to source according to different sign conventions and sign choices made in the course of the derivation.  It's inelegant because the rest of the theory of spinors can be developed without making such uncanonical choices. Both the uniqueness of the dimension of the irreducible Dirac representations (there are two of them in odd dimensions) and the existence of the Weyl spinors in even dimensions can be derived purely from the abstract properties of the Clifford algebra, no choices made, no transpose, adjoint or conjugate occuring. The (admittedly slightly subjective question) is: Is there a way to show in which dimensions Majorana spinors exist that neither requires an uncanonical choice of basis nor arbitrary choices of signs?  Some partial results:   In even dimensions, the Dirac representation is necessarily self-conjugate since it is the only irreducible representation of the Clifford algebra, so all that is left to show is that a conjugate-linear $\mathfrak{so}$-equivariant map on it squares to $\mathrm{id}_V$ and not to $-\mathrm{id}_V$. However, I can't seem to exhibit any particular equivariant map on it that one could simply check for its square. In odd dimensions, one first needs to figure out whether the two inequivalent Dirac representations are conjugate to each other or self-conjugate.     As further motivation that a clear proof using only canonical properties of the Clifford algebra itself is required, consider the confusing and contradictory claims in the literature:   Polchinski, "String Theory", Vol. 2, p.434: $\mathrm{SO}(d-1,1)$ has Majoranas for $d = 0,1,2,3,4 \mod 8$, corresponding to a special case of $p-q = d-1-1 = 6,7,0,1,2 \mod 8$. Fecko, "Differential Geometry and Lie Groups for Physicists", pp. 651: $\mathrm{Cliff}(p,q)$ has Majoranas for $p-q = 0,2\mod 8$.  This clearly conflicts with Polchinski's claims e.g. for $d=3$. Figueroa-O'Farrill, , pp. 18: We have Majoranas for $p-q = 0,6,7\mod 8$ and "symplectic Majoranas" for $p-q = 2,3,4\mod 8$.   Note that these results conflict simply by the number of possible $p-q$ regardless of whether I've correctly taken care of the differing conventions of whether $p$ or $q$ denotes timelike dimensions.
86944
There is an exercise in Humphreys's An Introduction to Lie Algebras and Representation Theory: "Any nilpotent Lie algebra contains a codimension 1 ideal".  The proof I am thinking of is the following.  Suppose the Lie algebra $L$ satisfies $L\neq[L,L]$. Then $\pi\colon L\to L/[L,L]$ is a projection onto an abelian algebra. Choose a codimension 1 subspace $I\subset L/[L,L]$. Then $I$ is an ideal and its preimage $\pi^{-1}(I)$ is necessarily an ideal in $L$, because any subspace containing the derived algebra is an ideal. Moreover, $\pi^{-1}(I)$ has codimension 1. $\square$  The only condition on $L$ is that $L\neq[L,L]$, which holds for solvable Lie algebras as well. Is this proof correct? How come Humphreys didn't set the exercise with "nilpotent" replaced by "solvable"?
62771
It is often said that the classical charge $Q$ becomes the quantum generator $X$ after quantization. Indeed this is certainly the case for simple examples of energy and momentum. But why should this be the case mathematically?  For clarity let's assume we are doing canonical quantization, so that Poisson brackets become commutators. I assume that the reason has something to do with the relationship between classical Hamiltonian mechanics and Schrodinger's equation. Perhaps there's a simple formulation of Noether's theorem in the classical Hamiltonian setting which makes the quantum analogy precisely clear?  Any hints or references would be much appreciated!  Mathematical Background  In classical mechanics a continuous transformation of the Lagrangian which leaves the action invariant is called a symmetry. It yields a conserved charge $Q$ according to Noether's theorem. $Q$ remains unchanged throughout the motion of the system.  In quantum mechanics a continuous transformation is effected through a representation of a Lie group $G$ on a Hilbert space of states. We insist that this representation is unitary or antiunitary so that probabilities are conserved.   A continuous transformation which preserves solutions of Schrodinger's equation is called a symmetry. It is easy to prove that this is equivalent to $[U,H] = 0$ for all $U$ representing the transformation, where $H$ is the Hamiltonian operator.  We can equivalently view a continuous transformation as the conjugation action of a unitary operator on the space of Hermitian observables of the theory  $$A \mapsto UAU^\dagger = g.A$$  where $g \in G$. This immediately yields a representation of the Lie algebra on the space of observables  $$A \mapsto [X,A] = \delta A$$  $$\textrm{where}\ \    X \in \mathfrak{g}\ \ \textrm{and} \ \  e^{iX} = U \ \  \textrm{and} \ \  e^{i\delta A} = g.A$$   $X$ is typically called a generator. Clearly if $U$ describes a symmetry then $X$ will be a conserved quantity in the time evolution of the quantum system.  Edit  I've had a thought that maybe it's related to the 'Hamiltonian vector fields' for functions on a symplectic manifold. Presumably after quantization these can be associated to the Lie algebra generators, acting on wavefunctions on the manifold. Does this sound right to anyone?
74598
I have just started learning Lie Groups and Algebra. Considering a flat 2-d plane if we want to translate a point from $(x,y)$ to $(x+a,y+b)$ then can we write it as :  $$   \left( \begin{array}{ccc} x+a  \\ y+a  \end{array} \right) =   \left( \begin{array}{ccc} x  \\ y  \end{array} \right) +    \left( \begin{array}{ccc} a  \\ b  \end{array} \right)$$  Now the set of all translations  $ T =   \left( \begin{array}{ccc} a  \\ b  \end{array} \right) $ form a two parameter lie group (I presume) with addition of column as the composition rule.  If that is so, how do I go about finding the generators of this transformation. I know the generators of translation are linear momenta in the corresponding directions. But I am not able to see this here.  PS:  In my course I have been taught that the generators are found by calculating the Taylor expansion of the group element about the Identity of the group. For instance, $\operatorname{SO}(2)$ group  $$ M =   \left( \begin{array}{cc} \cos \:\phi &amp; -\sin \:\phi \\ \sin \:\phi &amp;  \cos \:\phi \end{array} \right) $$ I obtain the generator by taking  $$ \frac{\partial M}{\partial \phi}\Bigg|_{\phi=0} =   \left( \begin{array}{cc} 0  &amp; -1 \\ 1 &amp;  0 \end{array} \right) $$  Now if I exponentiate this, I can obtain back the group element. My question how do I do this for Translation group.   EDIT :This edit is to summarise and get a view of the answers obtained.  Firstly, the vector representation of the translation group (for 2D) would in general have the form :  $$ \begin{pmatrix} 1 &amp; 0 &amp; a_x\\ 0 &amp; 1 &amp; a_y \\ 0 &amp; 0 &amp; 1 \end{pmatrix}\ $$ with generators (elements of Lie algebra)  $$  T_x =\begin{pmatrix} 0 &amp; 0 &amp; i\\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \end{pmatrix}\ , \;\;  T_y = \begin{pmatrix} 0 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; i \\ 0 &amp; 0 &amp; 0 \end{pmatrix}\  $$  Secondly, the scalar-field representation of the same is given by the differential operators $$ exp^{ i(a_x\frac{\partial}{\partial x}+ a_y\frac{\partial}{\partial y} )} $$ with generators  $$ T_x^s = i\frac{\partial}{\partial x},\;\;T_y^s = i\frac{\partial}{\partial y} $$  The Lie algebra is two-dimensional and abelian : $ [T_x,T_y] = 0$
115450
I have recently been reading Intro to Lie algebras and representation theory by Humphreys, and when I am finished I am interested in reading about Lie groups and Lie algebras and their applications to particle physics.     Is there a book that assumes basic knowledge of Lie algebras, and no knowledge of lie groups and particle physics/quantum mechanics?   I have seen Howard Georgi's book, but it assumes good knowledge of particle physics.   In addition, I do not know differential geometry.