One can show that the Lie algebra of $SO(3,1)$ is isomorphic to $SU(2)\times SU(2)$. And the representations of $SU(2)\times SU(2)$ exhausts all possible representations of $SO(3,1)$.   $\bullet$ Why does one consider, then, the irreducible representations of $SL(2,\mathbb{C})$?    $\bullet$ At the level of Lie algebra what is the relation between (i) $SL(2,\mathbb{C})$ and $SO(3,1)$ , (ii) $SU(2)\times SU(2)$ and $SL(2,\mathbb{C})$?  $\bullet$ I have found that, at the group level, $SO(3,1)$ is isomorphic not to $SU(2)\times SU(2)$ but to $SU(2)\times SU(2)/\mathbb{Z}_2$. Is this statement correct?
67335
This question is based on  problem II.3.1 in Anthony Zee's book Quantum Field Theory in a Nutshell      Show, by explicit calculation, that $(1/2,1/2)$ is the Lorentz Vector.    I see that the generators of SU(2) are the Pauli Matrices and the generators of SO(3,1)is a matrix composed of two Pauli Matrices along the diagonal. Is it always the case that the Direct Product of two groups is formed from the generators like this?   I ask this because I'm trying to write a Lorentz boost as two simultaneous quatertion rotations [unit quaternions rotations are isomorphic to SU(2)] and tranform between the two methods. Is this possible?      In other words, How do I construct the SU(2) representation of the Lorentz Group using the fact that $SU(2)\times SU(2) \sim SO(3,1)$?  Here is some background information:  Zee has shown that the algebra of the Lorentz group is formed from two separate $SU(2)$ algebras [$SO(3,1)$ is isomorphic to $SU(2)\times SU(2)$] because the Lorentz algebra satisfies:   $$\begin{align}[J_{+i},J_{+j}] &amp;= ie_{ijk}J_{k+} &amp; [J_{-i},J_{-j}] &amp;= ie_{ijk} J_{k-} &amp; [J_{+i},J_{-j}] &amp;= 0\end{align}$$  The representations of $SU(2)$ are labeled by $j=0,\frac{1}{2},1,\ldots$ so the $SU(2)\times SU(2)$ rep is labelled by $(j_+,j_-)$ with the $(1/2,1/2)$ being the Lorentz 4-vector because and each representation contains $(2j+1)$ elements so $(1/2,1/2)$ contains 4 elements.
15293
In John Fraleigh's book, A First Course In Abstract Algebra Exercises 15.7 and 15.11, one shows that  $$ (\mathbb{Z} \times \mathbb{Z})/ \langle (1,2)\rangle \cong \mathbb{Z} \times \mathbb{Z}_1 \cong \mathbb{Z} \ \ \ \ \  \mbox{ and  } \ \ \ \ \ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,2)\rangle \cong \mathbb{Z} \times \mathbb{Z}_2 $$  One does this with the first isomorphism theorem. With the same idea I proved for example that  $$ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,3)\rangle \cong \mathbb{Z} \times \mathbb{Z}_1 \cong \mathbb{Z} \ \ \ \ \ \mbox{ and }\ \ \ \ \ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,4)\rangle \cong \mathbb{Z} \times \mathbb{Z}_2 $$ So I conjectured that $(\mathbb{Z} \times \mathbb{Z})/ \langle (m,n)\rangle \cong \mathbb{Z} \times \mathbb{Z}_{k}$ where $k=\mathrm{gdc}(m,n)$. For the previous four cases, the homomorphism $\phi: \mathbb{Z}\times \mathbb{Z} \to \mathbb{Z} \times\mathbb{Z}_k$ given by $$ \phi(x,y)=\left(\frac{nx-my}{k}, \ x \ \ (\mathrm{mod} \ k) \right) $$ is surjective with kernel =$\langle (m,n)\rangle$. However, this is not the case when $(m,n)=(4,6)$. So, I cannot use what I did for the four cases to prove the general case.     What I want to know is if my conjecture is true. If so, how can I give a general homomorphism? If it is not true, how can I find $k$, such that $(\mathbb{Z} \times \mathbb{Z})/ \langle (m,n)\rangle \cong \mathbb{Z} \times \mathbb{Z}_k$?   Thanks in advance for any help/hint/comment!
34229
Let $GL_2(R)$ be the general linear group of $2\times2$ matrices and $SL_2(R)$ the special linear group of $2\times2$ matrices. Show that the commutator subgroup of $GL_2(R)$ is $SL_2(R)$.  I can show that the commutator subgroup is contained in $SL_2(R)$ as if $A,B \in GL_2(R)$ then $$\operatorname{det}(ABA^{-1}B^{-1})=\operatorname{det}(A)\operatorname{det}(B)\operatorname{det}(A^{-1})\operatorname{det}(B^{-1})=1.$$   But how can I show the reverse inclusion? That is, that $SL_2(R)$ is contained in the commutator subgroup of $GL_2(R)$.  Any help will be appreciated.
11375
Inspired by the , I am particularly looking into the exact sequences of SU(N), but I will like to loosen the conditions of the previous ,     Q1. $$1 \to A \to SU(N) \to B \to 1$$   where any of $A$,$B$ can contain finite group or discrete groups.    I am particularly interested in SU(5),     Q2. $$1 \to A \to SU(5) \to B \to 1$$    I wonder whether there is any example such that both $A$ and $B$ contains the same number of generators of continuous part of Lie groups (there can be an additional discrete finite group parts such as additional $\mathbb{Z_N}$)? (for example $SU(5)$ has 24 generators. Is there any example such that $A$ and $B$ have 12 generators for continuous part of Lie groups for each $A$, $B$.) How about  $$A=U(1) \times SU(2) \times SU(3)\;\;\; \text{or} \;\;\;A \supset U(1) \times SU(2) \times SU(3)?$$  Or whether there is any example such that $B$  $$B=U(1) \times SU(2) \times SU(3)\;\;\; \text{or} \;\;\;B \supseteq U(1) \times SU(2) \times SU(3)?$$ (there can be an additional discrete finite group parts for $B$ such as additional $\mathbb{Z_N}$.)
7537
I am trying to prove that the quotient group $SL_2(\mathbb Z_3)/Z(SL_2(\mathbb Z_3))$ is isomorphic to $A_4$. I could show that $SL_2(\mathbb Z_3)$ has $24$ elements (one can see this calculating all the elements by hand or simply by noticing that $GL_2(\mathbb Z_3)$ has $48$ elements and that $SL_2(\mathbb Z_3)$ is the kernel of the morphism $\det:GL_2(\mathbb Z_3) \to G_2$). The center of $SL_2(\mathbb Z_3)$ is $\left\{\begin{pmatrix}1&amp; 0\\0&amp; 1\end{pmatrix}, \begin{pmatrix}-1&amp; 0\\0&amp; -1\end{pmatrix}\right\}$. I got stuck trying to show these two groups are isomorphic. By Sylow theory I can say that $n_3=4$, where $n_p=$ number of $p$-Sylow subgroups. Then I can define the action by conjugation on $X$ the set of all $3$-Sylow subgroups. So I have a morphism $\phi:SL_2(\mathbb Z_3) \to S(X)$. It is clear that $Z(SL_2(\mathbb Z_3)) \subset \ker \phi$. I don't know what to do from here. I would appreciate suggestions or an alternative solution to this problem. Thanks in advance.
37210
I have been trying to solve a problem concerning the Heisenberg Lie group $H$. Show that there does not exist a faithful representation $\rho:H\to\text{GL}(2,\mathbb{R})$.  Any ideas about how to solve it would be very appreciated.
48202
It is well know that $SO(3)$ is a 3-dimensional Lie group. My question is  $SO(3)$ don't has 2-dimensional Lie subgroups?
83590
I am looking at the tensor product of $n$ spin halfs (fundamental of $SU(2)$): \begin{equation} \left(\frac{1}{2}\right)^n = \frac{1}{2} \times \frac{1}{2} \times ... \times \frac{1}{2} = \frac{n}{2} + ... \end{equation} For example, for $n = 3$ \begin{equation} \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{2} \times \left(1 + 0\right) = \left(\frac{3}{2}\right)_1 + \left(\frac{1}{2}\right)_1 + \left( \frac{1}{2} \right)_0 \end{equation} where lower indexes on the right hand side specify from which tensor product representation came from. My question is how do I read off if a certain term on the right is a symmetric, antisymmetric or a mixed symmetry representation of $SU(2)$? I came up with two ways of approaching the problem.  First is Young Tableaux. As far as I understand rows of Young Tableaux correspond to symmetric indexes while columns to antisymmetric ones. So, for example, a diagram with just one row (like one that corresponds to the highest spin $\frac{3}{2}$ from the example - three boxes in a row) describes a totally symmetric representation. A diagram that has one column (maximum 2 boxes since we are dealing with $SU(2)$) describes totally antisymmetric representation. Everything that has both rows and columns is a mixed symmetry representation. Following this logic it looks like only the highest representation in the product can be totally symmetric, singlet is antisymmetric and everything in between has mixed symmetry. So, for the example \begin{equation} \left(\frac{3}{2}\right)_1: \text{symmetric rep}\\ \left(\frac{1}{2}\right)_{0,1}: \text{mixed symmetry reps}\\ \end{equation} Is this is a correct logic?  Another way to think about this is the following. Let's build representation the way we do in QM starting from the highest one, when all the spins are up. Then all components of the highest multiplet can be generated by applying lowering operators to the highest state. So, all the relative signs in the expressions are &quot;+&quot; and thus the highest representation is always symmetric as 1. also suggests. Then all the other representations must be constructed by writing down expressions orthogonal to components of the highest multiplet, so they will necessary get some &quot;-&quot; signs and as a result must be either antisymmetric or have mixed symmetry, but can not be totally symmetric.   Is this correct logic? Is it true that only the highest representation in the product can be symmetric and all the others have mixed symmetry or antisymmetric?
68885
Modified question There was $sl(2,\mathbb R)$ used instead of $su(2)$ in the previous version. Thanks to MattE for pointing it out.  I have seen it claimed many times that $so(4,\mathbb R)=su(2)\times su(2)$. If $C$ is a Cartan subalgebra of $so(2n)$ then the Weyl group on $so(2n)$ is generated by permutations of $(x_1,...,x_n)\in  C$ and by negations of even number of $x_i$s. Hence, for $so(4)$, the orbit of $(x_1,x_2)\in C$ under the Weyl group action is $(x_1,x_2),  (-x_1,-x_2), (x_2,x_1), (-x_2,-x_1).$  The Weyl group of $su(2)\times su(2)$ is the product of Weyl groups of its two components. Hence, the orbit of $(x_1,x_2)$ in a Cartan subalgebra of $su(2)\times su(2)$ under its Weyl group action is $(x_1,x_2), (-x_1,x_2), (x_1,-x_2), (-x_1,-x_2).$  That looks like a contradiction! Where do I go wrong? You can of course reformulate this question for $SO(4,\mathbb R)=SU(2)\times SU(2)$ or $SO(4,\mathbb C)=SL(2,\mathbb C)\times SL(2,\mathbb C).$    Original question    I have seen it claimed many times that $SO(4,\mathbb R)=SL(2,\mathbb R)\times SL(2,\mathbb R),$ However, the orbit of $(x_1,x_2)$ in a max torus of $SO(4)$ under the Weyl group action is $(x_1,x_2), (x_1^{-1},x_2^{-1}), (x_2,x_1), (x_2^{-1},x_1^{-1}),$ while the orbit of $(x_1,x_2)$ in a max torus of $SL(2)\times SL(2)$ under the Weyl group action is $(x_1,x_2), (x_1^{-1},x_2), (x_1,x_2^{-1}), (x_1^{-1},x_2^{-1}).$  That looks like a contradiction! Where do I go wrong?
26055
I'm trying to prove the following. Let $\frak{g}$ and $\frak{h}$ be (semisimple) Lie algebras. Then every representation $d$ of $\frak{g}\oplus\frak{h}$ is the tensor product of representations $d^1$ on $\frak{g}$ and $d^2$ on $\frak{h}$; that is $d=d^1\otimes d^2$.  $d^1$ and $d^2$ are cannot be defined by restriction (see comment below). I don't know how to define them so that their tensor product must give back $d$. In particular how do I know that the vector space $V$ of $d$ decomposes appropriately for this to work?  Any hints would be much appreciated!
115829
I am a beginning learner of group theory.  Which of the following are isomorphic:$$\mathbb{Z_{24}}, D_{4}\times \mathbb{Z_{3}},A_{4}\times \mathbb{Z_{2}},\mathbb{Z_{2}}\times D_{6}, \mathbb{Z_{12}}\times \mathbb{Z_{2}},D_{12}, S_{4}$$  I know $\mathbb{Z_{24}}$ is cyclic and none of $\mathbb{Z_{12}}\times \mathbb{Z_{2}},D_{12}, S_{4}$ are cyclic since none has an element of order 24. I also know $D_{12}$ and $S_{4}$ are not isomorphic because $S_{4}$ has no element of order 12. Finally, I know $A_{4}$ and $D_{6}$ are also not isomorphic because $D_{6}$, unlike $A_{4}$, contains an element of order 6.  Other than this I don't know how to proceed. Any hints are appreciated. Please allow me to do the work.
32507
Show that (up to isomorphism) there is a unique Lie algebra over $F$ of dimension $3$ whose derived algebra has dimension $1$ and lies in $Z(L)$.   I think that I must construct a basis for $L$ satisfying such conditions. My first idea is to consider any canonical basis $x,y,h$ and calculate $[x,y],[x,h]$ and $[y,h]$. Then finding some relation between those brackets to define a new basis $x',y',h'$ such that whenever I compute the brackets of the new basis elements I will always lie in the subspace $[LL]$.  But unforunatly i don't know how. Any help would be much appreciated!
105655
I saw  and it states that the Lie algebra of the special linear group $SL_n(\mathbb C)$ is the group of traceless matrices $\mathfrak{sl}_n$.  I know the definition of a Lie algebra: given a matrix group $G \subseteq GL_n(\mathbb K)$ its Lie algebra is defined to be the tangent space at $I$.  From this definition it is immediately clear what dimension the Lie algebra has to have.  What is unclear to me is:      Assuming I don't already know the answer, how do I arrive at the insight that this tangent space is traceless matrices?   I skipped through Tapp's book in search of hints and found that the exponential map is a homomorphism from the Lie algebra into the Lie group. I also found a lemma that states that   $$ \det e^A = e^{\mathrm{trace}(A)}$$  That is unfortunately only half-useful: If I already know what the Lie algebra is then this lets me prove that the image of the Lie algebra under the homomorpism defined by the exponential map is indeed in $SL_n$.   So it's kind of not so useful. Surely there must be geometric insight?
41800
Hi I just start learning Lie algebra and there is one hw question I don't really understand how to do, hope somebody give me some hints. $L_1,L_2$ are Lie algebras. $L=\{(x_1,x_2):x_i \in L_i\}$. Lie bracket of $L$ is $$[(x_1,x_2),(y_1,y_2)]=([x_1,y_1],[x_2,y_2])$$ Call $L$ the direct sum of $L_1,L_2$. The question is to prove $gl(2,\mathbb C)$ is isomorphic to the direct sum of $sl(2,\mathbb C)$ with $\mathbb C$, the 1-dimensional complex abelian Lie algebra.  I have think about this for so long, but couldn't find a function that keeps the bracket structure. Can somebody help me with this. Thank you very much.
90454
I am slightly confused over the semantics of the definition of complete reducibility of a representation of a Lie algebra. Here's what I understand it to mean:  First, A representation $\phi:g \rightarrow gl(V)$ of a Lie algebra $g$ is irreducible if the vector space $V$ has no invariant subspaces.  Where we say a subspace $W$ of $V$ is invariant if it is invariant under $\phi(x)$ for all $x$ in $g$.  Now, it is said that a representation is completely reducible if it decomposes into a direct sum of irreducible representations (Wikipedia).  I struggle to understand what the last sentence means. What I think it means is that $V$ decomposes into a direct sum of vector spaces $V_i$ which are all irreducible (IE have no invariant subspaces).  This makes sense to me, but I also am not quite sure that I understand why it's not trivial. Should it not be the case that for any representation we will always be able to decompose $V$ into a direct sum of subspaces which are irreducible? I mean, the answer is obviously not (unless $g$ is semisimple), but I'm struggling with an intuitive understanding of that. Any help is appreciated.
105623
I have homework in a course on Lie groups, in which I must show that $(\pi_m,\mathbb C_m[x,y])$ are the only irreducible representations of finite dimension in SL(2,ℝ). Here $ C_m[x,y])$ is the vector space of homogeneous polynomials, and $\pi_m(g)$ acts via $\pi_m(g)(f(x,y))=f((x,y)\cdot g)$. It was not too complicated to calculate the characters, taking the basis $\{p_k:0\leq k \leq n\}$ with $p_k = x^ky^{n−k}$, since $\pi_m(exp(H))p_k=exp(d\pi_m(H))p_k=exp(2k−n)p_k$, but the question is how one could define an inner product for these matrices. But, I don't have the volume form of SL(2,ℝ) to define the integral $\langle f,g \rangle=\int_Gf(z)⋅\overline{g(z)}dz$
89718
Show $\operatorname{Aut}(C_2 \times C_2)$ is isomorphic to $D_6$ (the group with $x^3=1$, $y^2=1$ and $xy=yx^2$).  I'm not really sure how to express the elements of $\operatorname{Aut}(C_2 \times C_2)$. Would it be sufficient to show the elements of $\operatorname{Aut}(C_2 \times C_2)$, find their order and show they bijectively map to every element of $D_6$ and satisfy $xy=yx^2$?