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commutator_relations [2023/06/07 13:54] spencercommutator_relations [2023/06/07 13:55] (current) spencer
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 ==== Exterior and Interior products ==== ==== Exterior and Interior products ====
  
- - For forms $\alpha, \beta$, $e^*(\alpha)e(\beta) = e(e^*(\alpha)\beta) + (-1)^{|\alpha|} e(\beta) e^*(\alpha)$. Using $[x, y]_g$ for the graded commutator $xy - (-1)^{|x|}yx$, then $[e^*(\alpha), e(\beta)]_g = e(e^*(\alpha)\beta)$. In particular, for the 1-forms $\omega^i$ we have $e^*(\omega^i) e(\omega^j) + e(\omega^j) e^*(\omega^i) = 0$ for $i \neq j$, and is the identity otherwise.+    - For forms $\alpha, \beta$, $e^*(\alpha)e(\beta) = e(e^*(\alpha)\beta) + (-1)^{|\alpha|} e(\beta) e^*(\alpha)$. Using $[x, y]_g$ for the graded commutator $xy - (-1)^{|x|}yx$, then $[e^*(\alpha), e(\beta)]_g = e(e^*(\alpha)\beta)$. In particular, for the 1-forms $\omega^i$ we have $e^*(\omega^i) e(\omega^j) + e(\omega^j) e^*(\omega^i) = 0$ for $i \neq j$, and is the identity otherwise.
  
 ==== Connections ==== ==== Connections ====
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 This is the best way to figure these identities out on the fly and remember signs. This is the best way to figure these identities out on the fly and remember signs.
  
- - At the center of a nice frame, $\nabla_{\alpha} e(\omega^i) = e(\omega^i) \nabla_\alpha$; that is, $[\nabla, e(\omega^i)] = 0$. +    - At the center of a nice frame, $\nabla_{\alpha} e(\omega^i) = e(\omega^i) \nabla_\alpha$; that is, $[\nabla, e(\omega^i)] = 0$. 
- +    - Generally, $\nabla_\alpha e(\omega^i) = e(\omega^i) \nabla_\alpha + e^*(\nabla_\alpha \omega^i)$. 
- - Generally, $\nabla_\alpha e(\omega^i) = e(\omega^i) \nabla_\alpha + e^*(\nabla_\alpha \omega^i)$. +    - The dual version here is $[\nabla_\alpha, e^*(\omega^i)] = e^*(\nabla_\alpha \omega^i)$. 
- +    - The curvature tensor is $[\nabla_X, \nabla_Y] - \nabla_{[X,Y]}= F(X,Y)$. If $X, Y$ are coordinate vector fields (e.g. in a geodesic coordinate frame on the tangent bundle) then $[X,Y] = 0$ and $F(X,Y) = [\nabla_X, \nabla_Y]$.
- - The dual version here is $[\nabla_\alpha, e^*(\omega^i)] = e^*(\nabla_\alpha \omega^i)$. +
- - The curvature tensor is $[\nabla_X, \nabla_Y] - \nabla_{[X,Y]}= F(X,Y)$. If $X, Y$ are coordinate vector fields (e.g. in a geodesic coordinate frame on the tangent bundle) then $[X,Y] = 0$ and $F(X,Y) = [\nabla_X, \nabla_Y]$.+
commutator_relations.1686160468.txt.gz · Last modified: by spencer