In differential geometry, the slice theorem states:[1] given a manifold on which a Lie group acts as diffeomorphisms, for any in , the map extends to an invariant neighborhood of (viewed as a zero section) in so that it defines an equivariant diffeomorphism from the neighborhood to its image, which contains the orbit of .
![G/G_{x}\to M,\,[g]\mapsto g\cdot x](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e7fb97cc5b036a1eb73103b03f28eb8c4a8bd04)
The important application of the theorem is a proof of the fact that the quotient admits a manifold structure when is compact and the action is free.
In algebraic geometry, there is an analog of the slice theorem; it is called Luna's slice theorem.
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Proof of Projection Theorem
Transcription
- IN THIS QUESTION WE'RE ASKED TO EVALUATE THE DEFINITE INTEGRAL USING A GEOMETRIC FORMULA. SO IF A FUNCTION F OF X IS CONTINUOUS AND NON-NEGATIVE ON A CLOSED INTERVAL FROM A TO B THEN THE AREA UNDER THE FUNCTION AND ABOVE THE X-AXIS IS EQUAL TO THE DEFINITE INTEGRAL OF F OF X FROM A TO B. SO THE FIRST THING WE SHOULD NOTICE IS OUR CLOSED INTERVAL IS FROM -3 TO 3 GIVEN BY THE LIMITS OF INTEGRATION. NEXT, OUR FUNCTION F OF X IS EQUAL TO THE SQUARE ROOT OF 9 - X SQUARED. LET'S GO AHEAD AND SAY Y = THE SQUARE ROOT OF 9 - X SQUARED. THE NEXT THING WE SHOULD NOTICE IS THAT Y IS GOING TO BE NONNEGATIVE, AND THEREFORE THE VALUE OF THIS DEFINITE INTEGRAL WILL BE THE SAME AS THE AREA UNDER THIS FUNCTION ON THIS CLOSED INTERVAL. SO NOW WE WANT TO GRAPH THIS, AND TO HELP US RECOGNIZE HOW TO GRAPH THIS LET'S GO AHEAD AND SQUARE BOTH SIDES OF THE EQUATION. SO WE'LL SQUARE THE LEFT SIDE AND SQUARE THE RIGHT SIDE. SO THIS WOULD GIVE US Y SQUARED = THE SQUARING WILL UNDO THE SQUARE ROOT, SO WE HAVE Y SQUARED = 9 - X SQUARED. LET'S GO AHEAD AND ADD X SQUARED TO BOTH SIDES OF THE EQUATION. NOW WE SHOULD RECOGNIZE THAT THIS IS GOING TO BE A CIRCLE WITH A CENTER AT THE ORIGIN WITH A RADIUS OF 3, BUT REMEMBER SINCE Y IS NON-NEGATIVE WE'RE ONLY GOING TO SKETCH THE CIRCLE IN THE FIRST AND SECOND QUADRANTS WHERE Y IS POSITIVE. SO WE HAVE THE CENTER OF THE ORIGIN, AND SINCE THE RADIUS IS 3 WE CAN GO RIGHT 3 UNITS, UP 3 UNITS AND LEFT 3 UNITS TO FIND 3 POINTS ON THE CIRCLE. SO OUR FUNCTION WILL LOOK SOMETHING LIKE THIS. AND SINCE I DID SKETCH THIS ON SOME SOFTWARE, LET'S GO AHEAD AND USE THIS GRAPH HERE. AND AGAIN, BECAUSE THE FUNCTION IS NON-NEGATIVE ON THIS CLOSED INTERVAL THE AREA UNDER THIS FUNCTION AND ABOVE THE X-AXIS WILL BE EQUAL TO THE GIVEN DEFINITE INTEGRAL. SO IF WE CAN FIND THE AREA OF THE SHADED REGION WE CAN ALSO FIND THE VALUE OF THE DEFINITE INTEGRAL USING A GEOMETRIC FORMULA. AND SINCE THE AREA OF A CIRCLE IS GIVEN BY AREA = PI R SQUARED AND WE HAVE HALF A CIRCLE THIS AREA IS GOING TO BE EQUAL TO PI R SQUARED DIVIDED BY 2. NOTICE HOW IN THIS CASE AGAIN THE RADIUS IS EQUAL TO 3 WHICH MEANS THE DEFINITE INTEGRAL OF THE SQUARE ROOT OF 9 - X SQUARED FROM -3 TO 3 INTEGRATED WITH RESPECT TO X IS EQUAL TO PI x R SQUARED OR 3 SQUARED DIVIDED BY 2 WHICH IS 9 PI DIVIDED BY 2 SQUARE UNITS. I HOPE YOU FOUND THIS HELPFUL.
Idea of proof when G is compact
Since is compact, there exists an invariant metric; i.e., acts as isometries. One then adapts the usual proof of the existence of a tubular neighborhood using this metric.
See also
- Luna's slice theorem, an analogous result for reductive algebraic group actions on algebraic varieties
References
- ^ Audin 2004, Theorem I.2.1
External links
- On a proof of the existence of tubular neighborhoods
- Audin, Michèle (2004). Torus Actions on Symplectic Manifolds (in German). Birkhauser. doi:10.1007/978-3-0348-7960-6. ISBN 978-3-0348-7960-6. OCLC 863697782.
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This page was last edited on 15 January 2024, at 16:14