I realize that I don't update this blog very often, but so what? I'm the one paying for the domain, so I can post whenever I damn well please. And tonight, I feel like posting! I've been working with supersymmetry (SUSY) a lot lately, and several times people have asked me about what I mean when I refer to the Lorentz or Poincaré or supersymmetry algebra: what is an algebra? What do you mean it closes, or is isomorphic to a product algebra, or generates a Lie group? Well, I had a hard time describing it, so I'm going to try again tonight: An algebra is a module over a commutative ring equipped with a bilinear product \( [\cdot,\cdot]: A\times A \rightarrow A \). Okay...what does that mean? Essentially, an algebra extends the idea of a vector space by including some form of "vector product" that returns a vector of the original space. Indeed, the vector space \(\mathbb{R}^3\) equipped with the cross product forms just such an algebra over the field of real numbers. In physics, we often use a certain type of algebra, a Lie algebra \(\mathfrak{g}\), which has an additional product called the Lie bracket: it must be alternating, meaning \([x,x]=0\) for all \(x\in\mathfrak{g}\), and must satisfy the Jacobi identity \[ [x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0\] for all \(x,y,z\in\mathfrak{g}\). If the original algebra was equipped with a product, the the Lie bracket is identified with the commutator \([A,B]=A\cdot B-B\cdot A\). What does this have to do with physics? Well, it turns out that elements of a Lie algebra generate Lie groups, groups that can be infinitesimally varied, and hence are also manifolds. As such, for some Lie algebra element (called a generator of the Lie group) \(X\), we define \(e^{tX}\) to be an element of the Lie group \(G\) associated with the Lie algebra \(\mathfrak{g}\). An instructive example is the Lie algebra \(\mathfrak{su}(2)\), which is the complexification or universal (double) covering algebra of \(\mathfrak{so}(3)\), the angular momentum algebra. Its generators, \(J_x,J_y,J_z\) satisfy the algebra \[\left[J_x,J_y\right] = i\hbar J_z,\quad\left[J_y,J_z\right] = i\hbar J_x,\quad \left[J_z,J_x\right] = i\hbar J_y\] The corresponding Lie group is \(SU(2)\) or, if you prefer, \(SO(3)\), the group of orthogonal matrices of unit determinant. The elements of this algebra generate rotations in three dimensional space. This is a subalgebra of the Lorentz algebra \(\mathfrak{so}(3,1)\), whose generators obey the Lie bracket \[ [M^{\mu\nu},M^{\rho\sigma}]=i\left(\eta^{\nu\rho}M^{\mu\sigma}-\eta^{\mu\rho}M^{\nu\sigma}-\eta^{\nu\sigma}M^{\mu\rho}+\eta^{\mu\sigma}M^{\nu\rho}\right)\] Here, \(\eta^{\mu\nu}\) is the Minkowski metric of special relativity, which is the 4 by 4 identity matrix except the first entry is -1 (the time component). The generators are related to the \(J_i\) by \(J_i=\tfrac{1}{2}\epsilon_{ijk}M_{jk}\) and \(K_i=M_{0i}\), which generate the boosts in each direction. If we group them as \(A_i=\tfrac{1}{2}(J_i+iK_i)\) and \(B_i=\tfrac{1}{2}(J_i-iK_i)\), then we find that since \[\left[J_i,J_j\right]=i\epsilon_{ijk}J_k\quad\left[J_i,K_j\right]=i\epsilon_{ijk}K_k\quad \left[K_i,K_j\right]=-i\epsilon_{ijk}K_k\] we have \[\left[A_i,A_j\right]=i\epsilon_{ijk}A_k\quad \left[B_i,B_j\right]=i\epsilon_{ijk}B_k\quad \left[A_i,B_j\right]=0\] So, we get two copies of an \(\mathfrak{su}(2)\) algebra, meaning that locally, we have the isomorphism \(\mathfrak{so}(3,1)\cong\mathfrak{su}(2)\oplus\mathfrak{su}(2)\)! Furthermore, there is a homeomorphism \(\mathfrak{so}(3,1)\simeq\mathfrak{sl}(2,\mathbb{C})\). To see this, take a 4 vector and a corresponding 2 by 2 matrix: \[X=x_{\mu}e^{\mu}=(x_0,x_1,x_2,x_3)\quad \tilde{X}=x_{\mu}\sigma^{\mu}=\begin{pmatrix} x_0+x_3 & x_1-ix_2\\ x_1+ix_2 & x_0-x_3\end{pmatrix}\] where \(\sigma^{\mu}\) is the 4 vector of Pauli matrices (generating elements in the 2 dimensional spinor representation of \(\mathfrak{su}(2)\): \[\sigma^{\mu}=\left\{\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix},\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix},\begin{pmatrix} 0 & -i\\ i & 0\end{pmatrix},\begin{pmatrix} 1 & 0\\ 0 & -1\end{pmatrix}\right\}\] Transformations \(X\mapsto\Lambda X\) under \(SO(3,1)\) leaves the square \(|X|^2=x_0^2-x_1^2-x_2^2-x_3^2\) invariant, while the mapping \(\tilde{X}\mapsto N\tilde{X}N^{\dagger}\) with \(N\in SL(2,\mathbb{C})\) preserves the determinant \(\det\tilde{X}=x_0^2-x_1^2-x_2^2-x_3^2\). We see the homomorphism, as well as the fact that since \(N=\pm 1\) both  correspond to \(\Lambda=1\), the map is 2 to 1, and since \(SL(2,\mathbb{C})\) is simply connected, it is the universal covering group. Well, that's it for now...hopefully this was interesting to at least one person. If so, I'll continue with the Poincare group and its representations. Goodnight all! \(\k\)