An introductory course in differential geometry and the Atiyah-Singer index theorem

140 Pages·2006·1.72 MB·English

by  Paul Loya

An introductory course in di(cid:11)erential geometry
and the Atiyah-Singer index theorem
Paul Loya
Binghamton University, Binghamton NY 13902-6000
E-mail address: [email protected]
CHAPTER 1
What is the Atiyah-Singer index theorem?
1.1. The index of a linear map
Todescribetheindextheoremweofcourseneedtoknowwhat\index"means.
Section objectives: The student will be able to :::
(cid:15) de(cid:12)ne the index of a linear map.
(cid:15) compute the index for \simple" linear maps.
(cid:15) interpret the index analytically and topologically.
1.1.1. De(cid:12)nition and examples of the index. Let V and W be complex
vector spaces, not necessarily (cid:12)nite-dimensional, and let L : V W be a linear
!
map. Recall that the kernel (or null space) of L is de(cid:12)ned as
kerL:= v V Lv =0 :
f 2 j g
The cokernel of L is by de(cid:12)nition W modulo the image of L:
cokerL:=W=ImL;
therepresentsthespace\missedbyL"(notintheimageofL). NotethatbothkerL
andcokerLarevectorspaces. IfbothkerLandcokerLare(cid:12)nite-dimensional,then
we say that L is Fredholm. In a very rough sense, this means that L is \almost
bijective" in the sense that L is \almost injective" because L is injective except
on the (cid:12)nite-dimensional part kerL, and L is \almost surjective" because L only
a (cid:12)nite-dimensional part of W is missed by L. In particular, note that if L is an
isomorphism of the vector spaces V and W, then both the kernel and cokernel of
L are zero, so L is Fredholm.
When L is Fredholm, the dimensions of the vector space kerL and cokerL are
integers, and the index of L is de(cid:12)ned as the di(cid:11)erence:
indL:=dimkerL dimcokerL Z:
(cid:0) 2
In particular, the index of an isomorphism is zero. Here are some more examples.
Example 1.1. If V is (cid:12)nite-dimensional, observe that kerL V is automati-
(cid:26)
cally (cid:12)nite-dimensional. Also, if W is (cid:12)nite-dimensional, then cokerL = W=ImL
is automatically (cid:12)nite-dimensional. Thus, if both V and W are (cid:12)nite-dimensional,
then kerL and cokerL are (cid:12)nite-dimensional for any given linear map L:V W.
!
Inparticular,any linearmapbetween(cid:12)nite-dimensionalvectorspacesisFredholm;
in Theorem ? we compute the index of such a map.
Example 1.2. Before actually looking at Theorem ?, consider the example
1 2
L= :C2 C2:
2 4 !
(cid:18) (cid:19)
3
4 1. WHAT IS THE ATIYAH-SINGER INDEX THEOREM?
One can check that
2 1
kerL=span (cid:0) and ImL=span :
1 2
(cid:26)(cid:18) (cid:19)(cid:27) (cid:26)(cid:18) (cid:19)(cid:27)
Thus, dimkerL=1 and dimcokerL=dim(C2=ImL)=1, so indL=1 1=0.
Consider now L = Id : C2 C2. Then L is Fredholm with kerL(cid:0)= 0 and
cokerL=C2=ImL=0. Thus, in!dL=0 0=0.
Lastly, consider L = 0 : C2 C2 (t(cid:0)he zero map). Then L is Fredholm with
kerL = C2 and cokerL = C2=Im!L = C2, so indL = 2 2 = 0. Thus, we’ve seen
threemapsonC2,eachwithdi(cid:11)erentkernelsandcokern(cid:0)els,yetallhaveindexzero.
According to Theorem ?, any linear map L:C2 C2 will have index 0.
!
Example1.3. Inin(cid:12)nite-dimensions,linearmapsmayormaynotbeFredholm.
Consider V = W = C1(R), where C1(R) denotes the vector space of smooth, or
in(cid:12)nitely di(cid:11)erentiable, functions f :R C. Consider the linear map
!
L:C1(R) C1(R)
!
given by L(f) = sinx f; this map multiplies functions by sine. The map L
(cid:1)
is certainly a perfectly good (and nontrivial) linear map. Note that kerL = 0.
However, L is not Fredholm because, for example, the functions
1;x;x2;x3;:::;
de(cid:12)ne linearly independent elements of C1(R)=ImL as you can check. Hence,
cokerL=C1(R)=ImL is in(cid:12)nite-dimensional and therefore L is not Fredholm.
Example 1.4. Although not every linear map between in(cid:12)nite-dimensional
vector spaces is Fredholm there are many important examples. In this book the
mostimportant examplesare(cid:12)rstorderdi(cid:11)erentialoperators,exactlytheoperators
that appear in the Atiyah-Singer index theorem! Let us consider again V = W =
C1(R). Consider the di(cid:11)erentiation map
d
L= :C1(R) C1(R);
dx !
that is, L(f)=f0. Note that
kerL= f C1(R) f0 =0 = constant functions =C:
f 2 j g f g
Thus, dimkerL = 1. We claim that L is surjective. Indeed, given g C1(R),
x 2
de(cid:12)ne f(x) = g(t)dt. Then Lf = g, so L is surjective and dimcokerL = 0.
0
Hence, L is Fredholm and indL=1 0=1.
R (cid:0)
Example 1.5. One more example. Let us consider V = W = C1(S1), where
C1(S1) denotes the set of smooth functions on the unit circle S1, which are just
smooth functions f : R C that are periodic with period 2(cid:25). Denote by (cid:18) the
variable on S1 and consid!er the di(cid:11)erentiation map
d
L= :C1(R) C1(R)
d(cid:18) !
given by L(f)=f0. As before, we have
kerL= f C1(S1) f0 =0 = constant functions =C:
f 2 j g f g
Thus, dimkerL=1. In Problem 1 you will prove that dimcokerL=1 also. Thus,
L is Fredholm and indL=1 1=0.
(cid:0)
1.1. THE INDEX OF A LINEAR MAP 5
Later we shall see that the numbers 1 and 0 in the previous two examples are
exactly the Euler Characteristics of R and S1, respectively!
1.1.2. Theindextheanalytical. Beforeexplainingthetopologicalaspectof
the index we (cid:12)rst explain how the index is an \analytical object". Let L:V W
!
be a Fredholm linear map between two complex vector spaces and consider the
linear equation
(1.1) Lv =w for v V and w W:
2 2
Let v ;:::;v , where k = dimkerL, be a basis for kerL. Then assuming that
1 k
(1.1) holds for a v and w, by linearity, for arbitrary constants a ;:::;a C we
1 k
2
also have
L(v+a v + +a v )=w:
1 1 k k
(cid:1)(cid:1)(cid:1)
Thus, given that (1.1) holds for a v and w, in order to get a unique solution to
(1.1), we need to put constraints on the k =dimkerL constants a ;:::;a . Hence
1 k
we can think of
(1.2) dimkerL as the number of constraints needed for uniqueness in (1.1):
Let w ;:::;w W such that [w ];:::;[w ], is a basis for cokerL = W=ImL,
1 ‘ 1 ‘
2
where ‘ = dimcokerL and [ ] denotes equivalence class. Therefore, given any
w W, we can write
2
[w]=b [w ]+ +b [w ]
1 1 ‘ ‘
(cid:1)(cid:1)(cid:1)
for some constants b ;:::;b . Now observe that (1.1) just means that w ImW,
1 ‘
2
which is equivalent to [w]=0. Hence, given w W, there exists a solution v V
2 2
totheequation(1.1) ifandonlyifb =0;b =0;:::;b =0. Inotherwords, toget
1 2 ‘
existence to (1.1), we need to constraint the ‘=dimcokerL constants b ;:::;b to
1 ‘
vanish. Hence we can think of
(1.3) dimcokerL as the number of constraints needed for existence in (1.1):
In view of (1.2) and (1.3), we can think of
indL=dimkerL dimcokerL as the net number of constraints
(cid:0)
needed for existence and uniqueness in (1.1):
Thus, we can consider indL as \analytic" because existence and uniqueness of
solutions to (di(cid:11)erential) equations is a topic usually covered in an analysis course.
1.1.3. The index is topological. We now explain how the index is \topo-
logical". To do so, consider the following theorem.
Theorem 1.1. Let V and W be (cid:12)nite dimensional vector spaces. Then any
linear map L:V W is Fredholm, and
!
(1.4) indL = dimV dimW:
(cid:0)
Proof. Observethatifw ;:::;w areabasisforImW,thenwecancomplete
1 k
this set to a basis w ;:::;w ;u ;:::;u of W where ‘ = dimW dimImL. It
1 k 1 ‘
(cid:0)
follows that the equivalence classes [u ];:::;[u ] are a basis of W=ImL, so
1 ‘
dimW=ImL=‘=dimW dimImL:
(cid:0)
6 1. WHAT IS THE ATIYAH-SINGER INDEX THEOREM?
Figure 1.1. Computing the Euler Characteristic of R and S1.
Bythe\dimensiontheorem"or\ranktheorem"oflinearalgebra,wehavedimV =
dimkerL+dimImL. Hence,
indL=dimkerL dimW=ImL=dimkerL (dimW dimImL)
(cid:0) (cid:0) (cid:0)
=dimkerL+dimImL dimW
(cid:0)
=dimV dimW:
(cid:0)
Thus, indL=dimV dimW, which is a constant, regardless of L. (cid:3)
(cid:0)
Thistheoremistrivialtoprove,butprofoundtothinkabout: Notethattheleft-
handsideof (1.4)isananalytic object(inaccordancewithourpreviousdiscussion),
while the right-hand side of (1.4) is completely topological, involving the main
topological aspect of vector spaces, their dimensions. Thus, we can very loosely
interpret (1.4) as giving an equality8
\Analysis = Topology".
Another way the index is topological is that indL is an invariant of L in the
sense that it doesn’t change under (particular) deformations of L. We can see this
statement explicitly in Theorem 1.1 since the right-hand side of (1.4) is indepen-
dent of L; see Problem 2 for another example. Compare this invariance with the
correspondingsituationintopology: Topologicalaspectsofspaces(eg.connectivity,
compactness, etc.) are invariant under deformations (homeomorphisms).
Another topological aspect of the index can be seen in Examples 1.4 and 1.5.
LetusrecalltheEuler characteristic. Takeapenandmarkasmanydotsonthe
circleasyouwish,thencountthenumberdots(alsocalledvertices)andthenumber
of segments (also called edges) between adjacent dots, then subtract the numbers.
This di(cid:11)erence is by de(cid:12)nition the Euler characteristic of the circle, (cid:31)(S1), and it
does not depend on the number of dots you mark. In Figure (1.1) we have four
dots and four segments, so
(cid:31)(S1)=4 4=0:
(cid:0)
The Euler characteristic is topological in the sense that any space homeomorphic
tothecirclealsohasEulercharacteristiczero. Notethattheindexof d inExample
d(cid:18)
1.5 was also 0!
Tode(cid:12)netheEulercharacteristicofRwedothesameprocess: drawdotsonR
and count vertices and edges. However, we throw away the noncompact edges. For
example, in Figure 1.1 we have four vertices and three edges after throwing away
the noncompact edges, so
(cid:31)(R)=4 3=1:
(cid:0)
This is exactly the index of the (cid:12)rst order di(cid:11)erential operator d in Example 1.4!
dx
In conclusion, the index of (cid:12)rst order di(cid:11)erential operators (at least for the
examples we studied) is equal to a topological invariant of the underlying space!
ThisisessentiallytheAtiyah-Singerindextheorem, whichisvalidformoregeneral
1.2. HEAT KERNEL PROOF OF THE LINEAR ALGEBRA INDEX THEOREM 7
manifolds and (cid:12)rst order operators called Dirac operators. To give a rough idea
what the Atiyah-Singer theorem says, let V and W be certain smooth functions
on a space called a vector bundle. (More precisely, they are sections of a vector
bundle over a manifold; all this will be discussed in Chapter 2.) We shall study
operatorslike d and d alreadymentioned,andweshallconsidersuchanoperator
dx d(cid:18)
L:V W called a Dirac operator, named after the physicist Paul Dirac who (cid:12)rst
!
studied them; this will be done in Chapter 3. The Atiyah-Singer index formula
states that
indL= an expression involving speci(cid:12)c topological invariants of the space:
The topological invariants on the right are called characteristic classes. Thus, the
Atiyah-Singer index formula is a sense an equality:
\Analysis = Topology".
Exercises 1.1.
1. In the notation of Example 2, de(cid:12)ne
2(cid:25)
C1(S1)3f 7! f((cid:18))d(cid:18)2C;
Z0
ShowthatthismapiszeroonImL=Im d andhencede(cid:12)nesamapfromcokerL!C.
d(cid:18)
Prove that this map de(cid:12)nes an isomorphism cokerL(cid:24)=C. Hence dimcokerL=1.
2. In the notation of Example 1.4, de(cid:12)ne
d
L =a +b;
a;b dx
where a;b2C. Require a6=0 in order to keep L a (cid:12)rst-order di(cid:11)erential operator.
a;b
ShowthatthismapisFredholmandindL =1foralla;b2Cwitha6=0. Thisgives
a;b
an example showing that the index is stable under deformations.
3. In this exercise we see that indices of linear maps between (cid:12)xed in(cid:12)nite-dimensional
vector spaces can be arbitrary, in stark contrast to (cid:12)nite-dimensions. Let V = W =
C1(R).
(i) Find a linear map L:V !W that has index 2. Suggestion: Consider d2 .
dx2
(ii) Given any positive integer k, (cid:12)nd a linear map L:V !W with index k.
(iii) Given any negative integer k, (cid:12)nd a linear map L:V !W with index k.
1.2. Heat kernel proof of the linear algebra index theorem
The proof of Atiyah-Singer index formula that we will use is known as the
\heat kernel proof". To illustrate the main ingredients that go into proving the
Atiyah-Singer theorem, we will prove Theorem 1.1 using this heat kernel method.1
Before proceeding, keep Problem 1 at the end of this section in mind!
Section objectives: The student will be able to :::
(cid:15) understand the \heat kernel" proof of Theorem 1.1.
1Ofcourse,thelinearalgebraproofofTheorem1.1wasveryeasy,andtheproofwenowgive
is complicated in comparison, but the heat kernel proof generalizes to in(cid:12)nite dimensions while
theproofofTheorem1.1doesnot.
8 1. WHAT IS THE ATIYAH-SINGER INDEX THEOREM?
1.2.1. Theadjointmatrix. LetV andW be(cid:12)nite-dimensionalvectorspaces
and let L : V W be a linear map. For concreteness, we choose any basis of V
and W to iden!tify them with products of C. Thus, we can assume V = Cn and
W =Cm andwecanidentifyLwithanm ncomplexmatrix. LetL(cid:3) =Lt denote
(cid:2)
the adjoint, or conjugate transpose, matrix of L; note that L(cid:3) is an n m matrix.
(cid:2)
Properties of the adjoint include
(1.5) (AB)(cid:3) =B(cid:3)A(cid:3) and (A(cid:3))(cid:3) =A
for any matrices A and B (such that AB is de(cid:12)ned). Focusing on L, the adjoint
matrix has the property
(1.6) L(cid:3)w v =w Lv for all w Cm and v Cn;
(cid:1) (cid:1) 2 2
wherethedot denotesthestandarddot,orinner,product(onCn fortheleft-hand
side of (1.6) an(cid:1)d on Cm for the right-hand side of (1.6)). The adjoint matrix also
has the property
(1.7) kerL(cid:3) =cokerL;
(cid:24)
which you are to prove in Problem 2. Finally, note that since L is m n and L(cid:3) is
(cid:2)
n m, L(cid:3)L is an n n matrix and LL(cid:3) is an m m matrix.
(cid:2) (cid:2) (cid:2)
1.2.2. The heat operator. The(cid:12)rststepinprovingTheorem1.1istode(cid:12)ne
the \heat operators" of L(cid:3)L and LL(cid:3):
e(cid:0)tL(cid:3)L; e(cid:0)tLL(cid:3):
Here t is a real variable representing time, e(cid:0)tL(cid:3)L is an n n matrix (a linear map
on Cn), and e(cid:0)tLL(cid:3) is an m m matrix (a linear map o(cid:2)n Cm). We now discuss
(cid:2)
these operators more in depth. For concreteness, in the discussion we will focus on
e(cid:0)tL(cid:3)L although similar statements hold for e(cid:0)tLL(cid:3).
Theheatoperatore(cid:0)tL(cid:3)L iscommonlyfoundinordinarydi(cid:11)erentialequations
texts, cf. [3], where it is shown that given u Cn, u(t):=e(cid:0)tL(cid:3)Lu is the unique
0 0
2
solution to the initial value problem:
d
+L(cid:3)L u(t)=0; u(0)=u :
0
dt
(cid:16) (cid:17)
Theequation d +L(cid:3)L u(t)=0lookslikethe\heatequation"frommathematical
dt
physics that d(cid:16)escribes t(cid:17)he propagation of heat in a material. Therefore it’s not
unusual to call e(cid:0)tL(cid:3)L a heat operator.
There are two ways to de(cid:12)ne the heat operator. The (cid:12)rst way is by direct
exponentiation:
e(cid:0)tL(cid:3)L := 1 ((cid:0)t)k(L(cid:3)L)k:
k!
k=0
X
Onecanshowthatthissumisconvergentwithinthen nmatrices. Thesecondway
(cid:2)
to de(cid:12)ne the heat operator is via similar matrices. Note that L(cid:3)L is a self-adjoint
or Hermitian matrix, which means that it’s own adjoint: Using the properties in
(1.5) we see that
(L(cid:3)L)(cid:3) =L(cid:3)(L(cid:3))(cid:3) =L(cid:3)L:
1.2. HEAT KERNEL PROOF OF THE LINEAR ALGEBRA INDEX THEOREM 9
Being self-adjoint, L(cid:3)L is similar to a diagonal matrix with entries given by the
eigenvaluesofL(cid:3)L(thisissometimespartofthe\spectraltheorem"forself-adjoint
matrices). In other words, we can write
(cid:21)
1
(1.8) L(cid:3)L=U2 ... 3U(cid:0)1;
(cid:21)
6 p 7
6 0 7
6 7
4 5
for some matrix U, where the (cid:21) ’s represent the non-zero eigenvalues of L(cid:3)L, and
i
where 0 represents the 0 matrix. Note that the dimension of the 0 matrix is equal
to dimkerL(cid:3)L. We claim that dimkerL(cid:3)L = dimkerL. Indeed, if u kerL(cid:3)L,
2
then from property (1.6) we have
0=0 u=(L(cid:3)Lu) u=L(cid:3)(Lu) u=Lu Lu= Lu 2;
(cid:1) (cid:1) (cid:1) (cid:1) k k
where Lu denotes the norm of Lu. Thus, Lu=0 and so, kerL(cid:3)L kerL. Since
k k (cid:26)
kerL kerL(cid:3)L, it follows that kerL(cid:3)L = kerL, and so their dimensions are also
(cid:18)
equal. Now plugging (1.8) into the series e(cid:0)tL(cid:3)L := 1 ((cid:0)t)k(L(cid:3)L)k, one can
k=0 k!
show that
P
e(cid:0)t(cid:21)1
(1.9) e(cid:0)tL(cid:3)L =U2 ... 3U(cid:0)1;
e(cid:0)t(cid:21)p
6 7
6 Id 7
6 dimkerL 7
4 5
where Id is the identity matrix of dimension dimkerL(cid:3)L = dimkerL. If
dimkerL
you’re not conformable with the series de(cid:12)nition of e(cid:0)tL(cid:3)L, you can take (1.9) as
the de(cid:12)nition instead.
Switching L and L(cid:3), we can also get an expression
e(cid:0)t(cid:21)01
e(cid:0)tLL(cid:3) =U2 ... 3U(cid:0)1;
e(cid:0)t(cid:21)0q
6 7
66 IddimkerL(cid:3) 77
4 5
where (cid:21)0;:::;(cid:21)0 are the non-zero eigenvalues of L(cid:3). Now it is an important fact
1 q
that L(cid:3)L and LL(cid:3) have exactly the same non-zero eigenvalues. To verify this, let
u be a non-zero eigenvector of L(cid:3)L corresponding to some (cid:21) . Then observe that
i
(LL(cid:3))(Lu)=L(L(cid:3)Lu)=L((cid:21) u)=(cid:21) (Lu):
i i
Note that Lu = 0 since L(cid:3)Lu = (cid:21) u = 0. Thus, Lu is an eigenvector of LL(cid:3)
i
6 6
with eigenvalue (cid:21) . Hence, we’ve shown that L de(cid:12)nes a map from the eigenspace
i
of L(cid:3)L with eigenvalue (cid:21) into the eigenspace of LL(cid:3) with eigenvalue (cid:21) . One
i i
can check that L(cid:3)=(cid:21) is the inverse to this map. In conclusion, we’ve shows that
i
if (cid:21) is a nonzero eigenvalue of L(cid:3)L, then (cid:21) is also an eigenvalue of LL(cid:3) and
i i
the corresponding eigenspaces are isomorphic. One can also so the reverse: if
(cid:21)0 is a nonzero eigenvalue of LL(cid:3), then (cid:21)0 is also an eigenvalue of L(cid:3)L and the
i i
correspondingeigenspacesareisomorphic. Thisshowsthatthenonzeroeigenvalues
of L(cid:3)L and LL(cid:3) are the same. In particular, after reordering if necessary, we can
10 1. WHAT IS THE ATIYAH-SINGER INDEX THEOREM?
write
e(cid:0)t(cid:21)1
(1.10) e(cid:0)tLL(cid:3) =U2 ... 3U(cid:0)1:
e(cid:0)t(cid:21)p
6 7
66 IddimkerL(cid:3) 77
4 5
1.2.3. The McKean-Singer trick. Recall that the trace of a square matrix
is just the sum of the diagonal entries of the matrix. So, for example, if A = [a ]
ij
is a square matrix, then
TrA:= a
ii
i
X
The important property of the trace is that it is commutative in the sense that for
square matrices A and B of the same dimension,
TrAB =TrBA:
In particular, the trace is invariant under conjugation:
Tr(BAB(cid:0)1)=Tr(B(cid:0)1BA)=Tr(IdA)=Tr(A)
for any square matrix A and invertible matrix B of the same dimension as A.
Following McKean and Singer [2], we consider the function of t:
h(t):=Tr(e(cid:0)tL(cid:3)L) Tr(e(cid:0)tLL(cid:3)):
(cid:0)
This function has some amazing properties.
First,noticethatby(1.9)and(1.10)andtheconjugationinvariantofthetrace,
we have
h(t)= e(cid:0)t(cid:21)j +Tr(IddimkerL) e(cid:0)t(cid:21)j Tr(IddimkerL(cid:3))
(cid:0) (cid:0)
j j
X X
= e(cid:0)t(cid:21)j +dimkerL e(cid:0)t(cid:21)j dimkerL(cid:3)
(cid:0) (cid:0)
j j
X X
=dimkerL dimkerL(cid:3)
(cid:0)
=dimkerL dimcokerL;
(cid:0)
where we used (1.7). Thus, h(t) indL for all t.
Second, let us choose a parti(cid:17)cular t: Since e(cid:0)tL(cid:3)L and e(cid:0)tLL(cid:3) are the identity
matrices on Cn and Cm respectively at t = 0 (see the formulas (1.9) and (1.10)),
we have
indL=h(0)=Tr(Id ) Tr(Id )=n m=dimV dimW:
n m
(cid:0) (cid:0) (cid:0)
Thus,
indL=dimV dimW;
(cid:0)
and Theorem 1.1 is proved.
Exercises 1.2.
1. Dig out your linear algebra book and review properties of linear algebra you weren’t
familiar with in Section 1.2. These properties are fundamental and should be known!
2. Prove the equality in (1.7): kerL(cid:3) (cid:24)=cokerL. Suggestion: Observe that Cm =ImL(cid:8)
(ImL)? where(ImL)? istheorthogonalcomplementofImLinCm. Thus,cokerL=
Cm=ImL(cid:24)=(ImL)?. Now prove that kerL(cid:3) =(ImL)?.
CHAPTER 2
Manifolds and Riemannian geometry
2.1. Smooth manifolds
Intuitively a smooth manifold is a set which near each point \looks like" an
opensubsetofEuclideanspace;inthissectionwemakethisintuitivenotionprecise.
Section objectives: The student will be able to :::
(cid:15) de(cid:12)ne what a smooth manifold is.
(cid:15) verify that a set with an atlas is a manifold.
2.1.1. Smooth functions. We begin by discussing smooth functions. Let
Rn be open. A function f : R is said to be smooth or C1 if all partial
U (cid:18) U !
derivatives of all orders of f exist everywhere. In other words, if x ;:::;x are the
1 n
standard coordinates on Rn, then we require all (cid:12)rst partials to exist:
@f @f @f
; ;:::; ;
@x @x @x
1 2 n
all second partials to exist:
@2f @2f @2f
; ;:::; ;
@x2 @x @x @x @x
1 1 2 i j
all third, fourth, etc., to exist. We use the same terminology for complex-valued
functions, which we shall use a lot later on; thus, f : C is smooth or C1 if
U !
all partial derivatives of all orders of f exist.1 However, we shall concentrate on
real-valued functions for the (cid:12)rst section in this chapter.
Example 2.1. Any polynomial p : R is smooth. Here, a polynomial is
U !
just a (cid:12)nite sum of the form
p(x)=a+ a x + a x x + + a x x x ;
i i ij i j ij(cid:1)(cid:1)(cid:1)k i j k
(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)
where the coe(cid:14)cientsXa;a ;::: arXe real numbers. X
i
ThenextexamplesareusefulforconstructingpartitionsofunityinSection2.2.
Example 2.2. Here is an interesting example of a smooth function. De(cid:12)ne
e(cid:0)1=t if t>0;
f(t)=
(0 if t 0:
(cid:20)
Notice that f(t) is in(cid:12)nitely di(cid:11)erentiable for t = 0, so we just have to check that
6
f(t) is in(cid:12)nitely di(cid:11)erentiable at t=0. To this end, observe that
f(t) f(0) e(cid:0)1=t if t>0;
(cid:0) = t
t (0 if t 0:
(cid:20)
1Thisisequivalenttothesmoothnessoftherealandimaginarypartsoff.
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