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### Transcript of MATH 8210, FALL 2011 LECTURE usher/8210-notes.pdfMATH 8210, FALL 2011 LECTURE NOTES MIKE USHER 1....

• MATH 8210, FALL 2011 LECTURE NOTES

MIKE USHER

1. Multivariable calculus without coordinates

The objects of study in this course are what are called smooth manifolds. For the time being I wont givea precise definition of these (it will come later, or of course you can easily look it up), but for now suffice it tosay that these are topological spaces which locally resemble Euclidean space and in which, in particular, it ispossible to do something resembling calculus. The surface of the Earth is (to good approximation) an exampleof a two-dimensional smooth manifold. Of course, the Earth is not R2 but rather a closed surface (I was going tosay a sphere, but then it occurred to me that if one looks closely enough there are some rock formations whichcause the genus to be positive), yet locally it looks enough like R2 that it seems reasonable to speak for instanceof the directional derivatives of a function (the temperature, say) defined on the Earth.

So how can we formulate calculus in such spaces? Part of the definition will be that a manifold M will havean open cover {U| A} by sets equipped with homeomorphisms (charts) : U V where V Rn isopen. So we can try to do calculus on M by, roughly speaking, doing standard multivariable calculus in the opensets V and then transporting the constructions back to M by the maps (or their inverses). However, if m M,then m will typically belong to several of the sets U in the open cover of M, and one needs to make sure thatones constructions dont depend on which of the charts one is using. To compare between the th chart and theth chart, one needs to look at the transition function

1 : (U U) (U U).

This is a map between two open subsets of Rn, and part of the definition of a smooth manifold will ensure thatthe map is smooth (i.e., C) and invertible (with a smooth inverse), but there wont be any restrictions on what

1 other than that. So for example it doesnt make sense to take the partial derivative of a function on M

with respect to the first coordinate, since although we can differentiate a function on V with respect to the firstcoordinate, or we can do the same for a function on V, these operations wont be equivalent when we try to liftthem up to M using the maps , .

So this makes it important to understand how notions of multivariable calculus behave under the action ofdiffeomorphisms (i.e., smooth maps with smooth inverses) : U U where U and U are open subsets of Rn.You should think of the action of such a diffeomorphism as being the same as changing ones coordinate system,e.g. from Cartesian coordinates to polar coordinates. In particular I want to first discuss various notions of whata tangent vector at a point p U is. (And well later generalize this to the notion of a tangent vector at a pointin a smooth manifold.) Visually youre supposed to think of a tangent vector at p as being a little arrow whosebase is at p, pointing in a possible direction of motion from p. The set of these tangent vectors will form a vectorspace called the tangent space to U at p and denoted TpU. Ill give three characterizations, from most concreteto most abstract.

(1) The way to describe this notion that is used in undergraduate multivariable calculus courses is just to saythat a tangent vector v at p U is (or is represented by) an n-tuple of numbers (v1, . . . , vn) Rn. Onecan then draw the vector whose base is at p and whose first coordinate is v1, second coordinate is v2, andso on. (In somewhat more sophisticated language, the standard Cartesian coordinates on Rn determine abasis {e1, . . . , en} of unit vectors, and one has v =

viei.)

1

• 2 MIKE USHER

This characterization is very good for computational purposes, but when one is interested in howtangent vectors behave under coordinate changes : U U it has some disadvantages. The tangentvector v = (v1, . . . , vn) TpU should correspond under the coordinate change to a tangent vectorv T(p)U at (p). Perhaps youve learned how this correspondence works: one constructs theJacobian matrix at p of the map (with (i, j) entry given by i

x jwhere i is the ith component of ),

and then the coordinates of v are obtained by multiplying the Jacobian matrix by the vector consistingof the components of v. This is a manageable computation, but it may not be very conceptually clearfrom this discussion whats going on here. In particular if we then want to say what a tangent vectorto a point m on a smooth manifold is wed have to say something like an n-tuple of numbers for eachchart containing m, such that the n-tuples for different charts are related by the Jacobians of the transitionfunctions, which is much more opaque and less natural-sounding than it really should be.

(2) A more natural characterization of tangent vectors is the following. The idea is that the tangent spaceTpU consists of all possible velocities of curves passing through p. If p U, consider all C paths : (, ) U (for some > 0) such that (0) = p. I would like to declare two of these to beequivalent if they have the same velocity, i.e., 1 2 iff 1(0) =

2(0) (or equivalently, and maybe

less circularly, 1 2 if limt01(t)2(t)

t = 0). Then simply define a tangent vector at p to be anequivalence class [] of C arcs through p (and so TpU is just the set of equivalence classes). The waythis behaves under coordinate changes is extremely simple, since Im not using coordinates to define thenotion: a tangent vector v TpU has the form v = [] for some , and the corresponding tangent vectorv T(p)U is just [ ]. Well see later that this adapts to general smooth manifolds very simply anddirectlya tangent vector at a point on a smooth manifold will just be a suitable equivalence class ofcurves passing through that point.

The one disadvantage of this characterization is that its not so intuitively obvious how to do algebraicoperations (like addition of tangent vectors) on equivalence classes of curves through a point (thoughyou can make a suitable definition if you put your mind to it).

It shouldnt be hard to construct a natural correspondence between tangent vectors in this sense andtangent vectors in the sense of Definition (1) above, but again, the advantage of thinking about it thisway is that its less coordinate-dependent.

(3) Now for a characterization of tangent vectors that you almost certainly would not have thought of. Toattempt to motivate it, note that a given tangent vector v TpU gives you the ability to differentiatesmooth functions f : U R at pnamely you take the directional derivative at p:

(Dv f )(p) = limt0

f (p + tv) f (p)t

.

So we will define a tangent vector at p to be a way of differentiating functions defined near p, i.e., wewill abstract some relevant properties of the operation of taking a directional derivative, and then definea tangent vector to be one of these operations.

To do this, first consider pairs ( f ,V) where V is an open neighborhood of p and f : V R isC, and declare two such pairs ( f ,V) and (g,W) to be equivalent if there is a smaller neighborhoodZ V W of p such that f |Z = g|Z . Let Op be the set of equivalence classes. Since we can set,for instance [ f ,V] [g,W] = [ f g,V W], Op is easily seen to be a commutative R-algebra (i.e., it isboth a commutative ring and a vector space over R, with appropriately compatible operations), calledthe algebra of germs of functions at p. Ill tend to denote a germ by just f rather than [ f ,V]; itis to be understood that f is defined not necessarily throughout U but rather on some (varying) openneighborhood of p. Of course one always has a well-defined value f (p) for f Op.

A tangent vector at p will then be defined to be a derivation v : Op R, i.e. v is to satisfy (R-linearity) v(c f + g) = cv( f ) + v(g) for c R and f , g Op (Leibniz rule) v( f g) = f (p)v(g) + g(p)v( f ) for f , g Op.

• MATH 8210, FALL 2011 LECTURE NOTES 3

Its standard that the directional derivative operations Dv alluded to above satisfy these properties. Itsnot obvious that, conversely, any derivation on Op is given by a directional derivative in some direction,but well prove this shortly.

Like the characterization of tangent vectors as equivalence classes curves, this formulation is com-pletely coordinate free, making it easy to extend the definition to manifolds when the time comes. Unlikethe situation with curve characterization, though, its quite obvious that derivations form a vector space,which is another advantage.

To see how this notion behaves under diffeomorphisms (or indeed under more general smooth maps) : U U, if v TpU (i.e., if v is a derivation on Op), we need to construct a derivation v on O(p).Well, if f O(p) (really we should write [ f ,V]), so f is a smooth function defined near (p), then f will be a smooth function defined near p (specifically, it will be defined on the open set 1(V) aroundp), and so we can define

(v)( f ) = v( f )

So as with the curve formulation, its quite simple to see how derivations transform under coordinatechanges.

Among the three above characterizations of tangent vectors, it should be clear that (1) is equivalent to (2),under the correspondence which assigns to an equivalence class of curves [] the vector (0) (expressed incoordinates using the standard basis for Rn). We now set about proving that (1) and (3) are also equivalent.Let TpU denote the space of tangent vectors as given by formulation (1) (i.e., as elements of Rn) and (for themoment) TpU that given by (3) (i.e., as derivations). Write the coordinates of p U Rn as (p1, . . . , pn). Nowwe have a linear map : TpU TpU given by

(v1, . . . , vn) =n

i=1

vi

xi,

i.e., sends a vector (in the undergraduate multivariable calculus sense) to the operation