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Let $Q$ be a smooth manifold, and let $G$ be a Lie group which acts smoothly on $Q$ on the left.

  • Question 1: does the group $G$ act naturally on the tangent bundle $TQ \to Q$?

My motivation here is $Q = \mathbb R^d$ with group of symmetries the Euclidean group $G = \operatorname{Euc}(d) \cong \mathbb R^d \rtimes O(d)$. In this case, $G$ admits an equivariant action on the tangent bundle, where translations act by translating the underlying space and ignoring tangent spaces, and rotations act by rotating both the space and the tangent spaces. I am wondering if this generalizes nicely, or if this lifted action is special to Euclidean space.

Next, consider the frame bundle $GL(TQ) \to TQ$ as a principal $GL(d)$-bundle over the tangent bundle. Note that in local coordinates, $GL(TQ)$ looks like $TQ \times GL(d)$, and there is a natural right action of $GL(d)$ on the frame bundle.

  • Question 2: does the group $G$ act naturally on the frame bundle $GL(TQ) \to TQ$?

If the answer to Question 1 is yes, then the answer to Question 2 should also be yes, since we would just act on each of the vectors comprising a frame in the "obvious" way.

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If the action is smooth then yes, the action on $TQ$ is just the derivative of the action on $Q$. Is this your question, are there non-smoothable actions? –  Ryan Budney Mar 12 '13 at 22:38
    
@Ryan Budney, smooth actions are all I am concerned with; I edited the post. Thanks for the quick reply. Could you add a few more details and post that as an answer? –  Tom LaGatta Mar 12 '13 at 23:19
    
In fact, Ryan answered both questions. –  Misha Mar 12 '13 at 23:41
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1 Answer

up vote 3 down vote accepted

The chain rule for manifolds states that if $f : M \to N$ and $g : N \to Q$ are smooth, then the derivative $Df : TM \to TN$ and $Dg : TN \to TQ$ satisfy $$D(g \circ f) = Dg \circ Df$$

So if $g \in G$, let the action of $g$ on $Q$ be denoted $L_g : Q \to Q$. So $L_g \circ L_h = L_{gh}$, therefore $D(L_{gh}) = D(L_g) \circ D(L_h)$. So the action of $G$ on $TQ$ is the map $G \times TQ \to TQ$ given by $(g,v) \longmapsto D(L_g)(v)$.

For Q2, just lift the action on the tangent bundle to the frame bundle, i.e. $(g,(v_1,\cdots,v_n)) \longmapsto (D(L_g)(v_1), \cdots, D(L_g)(v_n))$.

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Thank you, @Ryan! –  Tom LaGatta Mar 12 '13 at 23:57
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