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--- commutator_relations [2002/01/10 22:48] – created spencer
+++ commutator_relations [2023/06/07 13:55] (current) – spencer
@@ Line 3: / Line 3: @@
 ==== 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 ====
- - At the center of a nice frame, $\nabla_{\alpha} e(\omega^i) = e(\omega^i) \nabla_\alpha$; that is, $[\nabla, e(\omega^i)] = 0$.
+The point of all these identities is the maxim of 'apply Leibniz everywhere possible'.
+For example, to compute the value of the connection on a 1-form $\alpha$, we pick an arbitrary vector $v$ and vector field $X$ to get
+$$\nabla_X (\alpha(v)) = (\nabla_X \alpha)(v) + \alpha (\nabla_X v); $$
+then use that $\nabla = d$ on functions to rewrite as
+$$(\nabla_X \alpha)(v) = X(\alpha(v)) - \alpha(\nabla_X v).$$
+This is the best way to figure these identities out on the fly and remember signs.
- - Generally, $\nabla_\alpha e(\omega^i) = e(\omega^i) \nabla_\alpha + e^*(\nabla_\alpha \omega^i)$.
+    - 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, $\nabla_\alpha e(\omega^i) = e(\omega^i) \nabla_\alpha + e^*(\nabla_\alpha \omega^i)$.
- - 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 $F(X,Y) = [\nabla_X, \nabla_Y]$.