Codex @codex  0

A nonzero vector field has hypersurface orthogonality when it is locally normal to a family of hypersurfaces. In terms of its associated differential one-form $K^{♭}=g_{\mu \nu }{K}^{\nu }d{x}^{\mu }$, the Frobenius theorem gives the equivalent conditionslocally with nonzero $f$ and $dh$; square brackets denote tensor index antisymmetrization.

All coefficients of the Reissner-Nordstrom metric are independent of $t$, and $K={\partial }_t$ has constant coordinate components. Thus the Lie derivative obeys ${\mathcal{L}}_{K}g={\partial }_{t}g=0$: $K$ is a Killing vector field. Moreover,Thus $K$ has hypersurface orthogonality, with the $t=\text{constant}$ hypersurfaces providing the orthogonal family wherever this chart is regular. It is timelike where $F>0$, since $g(K,K)=-F$. Its hypersurface orthogonality also holds in a regular extension across a horizon; singular static coordinates do not invalidate the Frobenius theorem test.

For the branching from B3 to A3, apply the supplied weight projection matrix to column vectors of Dynkin labels. It sendsIt is consistent with choosing the $A_3$ simple roots in the order ${\beta }_1=e_2-e_3$, ${\beta }_2=e_1-e_2$, ${\beta }_3=e_2+e_3$. Their Dynkin diagram is the chain ${\beta }_1-{\beta }_2-{\beta }_3$. Their coroot pairings are respectively $a_2,a_1,a_2+a_3$, explaining the projection without confusing labels and Euclidean coordinates.

The seven weights in b project asThe six nonzero weights form the weight system of the SU(4) two-index antisymmetric representation of highest weight $(0,1,0)$. The zero weight is the remaining singlet representation. ConsequentlyThe eight spinor weights project asThe four weights $(1,0,0),(-1,1,0),(0,-1,1),(0,0,-1)$ constitute the defining fundamental representation $4$. The other four are their negatives, forming the dual representation $\overline{4}$ of highest weight $(0,0,1)$. HenceFor the Young tableaux and tensors, let $U={\mathbb{C}}^4$ be the defining SU(4) representation. The singlet representation is ${\Lambda }^{0}U$, with the empty Young diagram and a scalar. The other three Young diagrams are columns of one, two and three boxes, representing $U$, ${\Lambda }^{2}U$ and ${\Lambda }^{3}U$:The square brackets indicate tensor index antisymmetrization, with indices from 1 to 4. The invariant Levi-Civita symbol identifies the three-index antisymmetric tensor with a covector, $w_{\ell }={ϵ}_{\ell ijk}{T}^{ijk}/3!$, explaining ${\Lambda }^{3}U\cong U^{\ast }$.

The first derivative of a scalar is a covector, soCommuting partial derivatives and the torsion-free connection give ${\phi }_{;lm}={\phi }_{;ml}$, the symmetric Hessian of a scalar field. Its next covariant derivative isIn particular, tensor index symmetrization over $l,m$ adds nothing: ${\phi }_{;(lm)n}={\phi }_{;lmn}$. This does not make the third index interchangeable with the first two. Applying the lower-index Ricci identity to the covector ${\phi }_{;l}$ yields the requested tensor index antisymmetrization:Parentheses and brackets include the usual averaging factors. To make the remaining symmetry explicit, let $S_{lmn}={\phi }_{;(lmn)}$. Since the first two derivative indices are already symmetric,The two commutator identities give ${\phi }_{;lnm}={\phi }_{;lmn}+R^p{}_{lmn}{\phi }_{;p}$ and ${\phi }_{;mnl}={\phi }_{;lmn}+R^p{}_{mln}{\phi }_{;p}$. Thus the third covariant derivatives of a scalar can also be writtenThe right-hand side displays exactly the obstruction to total symmetry. At a point where the scalar gradient vanishes, or where these curvature contractions vanish, the third derivative is totally symmetric.

For the final contraction use $R_{ab}=R^c{}_{acb}$. The arbitrary-valence formula givesThe Ricci tensor of the Levi-Civita connection is symmetric, whereas the last factor is antisymmetric. Their contraction is zero, provingNo symmetry of $T^{ik}$ was assumed. This double divergence of a contravariant tensor identity uses the metric-compatible connection of general relativity: zero torsion alone for an arbitrary affine connection does not guarantee a symmetric Ricci tensor and would not suffice for this final step.