MATH 250B: COMPLEX ALGEBRAIC GEOMETRY 19 3. Kahler¨ manifolds AK¨ahlermanifoldisacomplexmanifoldM with a K¨ahler form ! which is closed (d! = 0). A K¨ahler form is equivalent to a Hermitian metric h.Wedefinetheseand show how they are related on a single vector space V ,thenonthetangentbundleofM. (However, on a single vector space, it doesn’t make sense to talk about closed forms.) 3.1. K¨ahler forms. We started with the basic concept of a K¨ahler form. Suppose that n V is a vector space over C: V ⇠= C and WR = Hom(V,R), W = W C = W 1,0 W 0,1. C R ⌦ 1,0 1,1 Recall that WC = HomR(V,C)andW = HomC(V,C) WC. Now take W = W 1,0 W 0,1.Wewanttolookat ⇢ ⌦ W 1,1 = W 1,1 2W . R \^ R 1,1 Thus elements of WR are alternating real forms of type (1, 1). Example 3.1.1. The basic example is V = Cn, i i ! = dz dz = (dx + idy ) (dx idy )= dx dy . 2 j ^ j 2 j j ^ j j j ^ j j X X X n Since a b = a b b a,forz = x + iy, z0 = x0 + iy0 C this form is ^ ⌦ ⌦ 2 !(z,z0)= (x y0 y x0 ). j j j j j X 1,1 2 The first form of ! shows that it lies in W .Thelastformshowsitisin WR.The i ^ scalar 2 is needed to make the form real. All K¨ahler forms will be equivalent to these. 1,1 Lemma 3.1.2. ! WR if and only if ! : V V R is a skew symmetric R-bilinear form so that 2 ⇥ ! (3.1) !(Iu,Iv)=!(u, v) for all u, v V . 2 Proof. First note that the condition !(Iu,Iv)=!(u, v)isequivalenttothecondition (3.2) !(u, Iv)+!(Iu,v)=0 2 1,1 since I = 1. By definition, WR is the set of skew-symmetric forms ! on V so that 1,1 !C = ! C lies in W . This is equivalent to the condition that !C vanishes on pairs of vectors⌦ from V 1,0 or from V 0,1.ButV 1,0 is the set of all vectors of the form u˜ = u iIu where u V and 2 !(˜u, v˜)=!(u iIu, v iIv) = !(u, v) !(Iu,Iv) i(!(u, Iv)+!(Iu,v)) which is zero if and only if (3.1) and (3.2) hold. ⇤ 20 MATH 250B: COMPLEX ALGEBRAIC GEOMETRY Note that (3.2) implies that g(u, v):=!(u, Iv) is a symmetric bilinear pairing g : V V R since ⇥ ! g(v, u)=!(v, Iu)= !(Iu,v)=!(u, Iv)=g(u, v) Definition 3.1.3. A hermitian form on a complex vector space V is defined to be a map h : V V C so that ⇥ ! (1) h(u, v)isC-linear in u (2) h(u, v)isC-antilinear in v (3) h(v, u)=h(u, v) Note that (3) implies that h(v, v) R.Theformh is said to be positive definite if h(v, v) > 0forallv =0.Apositivedefinitehermitianformon2 V is also called a Hermitian metric on 6V . Proposition 3.1.4. There is a 1 1 correspondence between hermitian forms h on V 1,1 and forms ! WR given by 2 ! = h. = Furthermore, h(u, v)=g(u, v) i!(u, v) where g : V V R is given by g(u, v)=!(u, Iv). ⇥ ! Proof. For the second part, h(u, Iv)=g(u, Iv) i!(u, Iv). Since h(u, v)isconjugate linear in v, h(u, Iv)= ih(u, v)= !(u, v) ig(u, v). Comparing complex parts gives g(u, v)=!(u, Iv). ⇤ Example 3.1.5. The standard positive definite hermitian form on Cn is:

h(z,z0)= zjzj0 = (xj + iyj)(xj0 + iyj0 ) X X = (x x0 + y y0 ) i (x y0 y x0 ) j j j j j j j j

X g(z,z0) X !(z,z0) | {z } | {z }1,1 Definition 3.1.6. A K¨ahler form on V is a form ! WR whose corresponding hermitian form h is positive definite. In particular, K¨ahler2 forms are nondegenerate. 3.2. K¨ahler metrics. Suppose that (M,I) is an almost complex manifold. Then a Hermitian metric h on M is a Hermitian metric hx on the tangent space TM,x at each point which varies smoothly with x M. Associated to h we have: 2 ! = h = 1,1 which is a 2-form on M which is also in ⌦M which is equivalent to the equation !(Iu,Iv)=!(u, v) for any two vectors u, v TM,x at any point x M.ByDefinition3.1.6,! is a K¨ahler form. However we usually2 want ! to be closed.2 We say that the Hermitian metric h is a K¨ahler metric if the corresponding K¨ahler form ! is closed. MATH 250B: COMPLEX ALGEBRAIC GEOMETRY 21 Proposition 3.2.1. The real part of a Hermitian metric h is a Riemannian metric g on M which is also invariant under I: g(u, v)=!(u, Iv)=g(Iu,Iv) Proof. We know that g is a symmetric real form. If v =0 T is a nonzero vector, 6 2 M,x h(v, v)=h(v, v) is a positive real number. So, g(v, v)=h(v, v) > 0 So, g is a Riemannian metric on M. ⇤ Note that M is oriented since any complex vector space has a natural real orientation. Theorem 3.2.2. Given a Hermitian metric h on the complex manifold M, the volume form on M associated to the Riemannian metric g = h is equal to !n/n!. < To prove this, we need the matrices for general h, ! in local coordinates: n z =(z1, ,zn):U, C ,zj = xj + iyj ··· ! 1,0 0,1 and z =(z1, , zn)centeredatx0 U.Thendzj,dzj form bases for ⌦ , ⌦ . ··· 2 M,x0 M,x0 These are W 1,0,W0,1 for V = T .Soh ⌦1,1 is given by M,x0 2 M h = ↵ dz dz ij i ⌦ j 0 where ↵ij : M C (notation: ↵ij ⌦X ). These functions are given by: ! 2 M,C @ @ ↵ = h , . ij @x @x ✓ i j ◆ Since h(u, v)=h(v, u), ↵ = ↵ .Since z = i (z z), we have: ji ij = 2 i i ! = h = (↵ dz dz ↵ dz dz )= ↵ dz dz . = 2 ij i ⌦ j ji i ⌦ j 2 ij i ^ j X X Proof of Theorem 3.2.2. By a C-linear change of the coordinates z =(z1, ,zn), we @ ··· can arrange for to be ortho-normal at the point x0.Inotherwords, @xi @ @ ↵ (x )=h , = ij 0 x0 @x @x ij ✓ i j ◆ Then, at x0, i ! = dz dz = dx dy x0 2 j ^ j j ^ j by Example 3.1.1. So, X X !n = dx dy dx dy dx dy ⇡(1) ^ ⇡(1) ^ ⇡(2) ^ ⇡(2) ^···^ ⇡(n) ^ ⇡(n) ⇡ Sn X2 which is n!timesthevolumeformdx1 dy1 dxn dyn at the point x0.Sincethis hold at every point x M, !n/n!isthevolumeformateachpoint.^ ^···^ ^ 0 2 ⇤ 22 MATH 250B: COMPLEX ALGEBRAIC GEOMETRY 3.3. K¨ahler manifolds. Definition 3.3.1. A K¨ahlermanifold is a complex manifold with a K¨ahler metric which is, by definition, a Hermitian metric h so that ! is closed (d! =0). Corollary 3.3.2. On a compact K¨ahlermanifold M, for every 1 k n, the form !k is closed but not exact. I.e., [!k] =0 H2k(M).   6 2 Proof. !k is clearly closed: k k 1 d! = k! d! =0 If !k = d↵ then n k n k n d(↵ ! )=d↵ ! = ! ^ ^ which is impossible since the volume form is nonzero in H2n(M)whenM is an oriented compact manifold. ⇤ Corollary 3.3.3. A compact complex k-submanifold N of a K¨ahlermanifold M cannot be the boundary of a (real) submanifold of M. Proof. This follows immediately from Stokes’ Theorem. If N = @W then:

d!k =0= !k = vol N ZW ZN acontradiction,since!k is the volume form on N. ⇤

3.4. Connections. Given a real C1 k-dimensional vector bundle E on a real manifold X we want to take the derivative of a section of E.Thisisgivenbyaconnection.Recall that a connection on E is a linear map r 0 1 : A (E) A (E)=(T ⇤ E) r ! X ⌦R satisfying the Leibnitz equation (f)=df + f r ⌦ r 0 0 for all f : X R (f ⌦X ), E = A (E)=C1(E). Thus, takes! a section2 of2E and gives () A1(E)whichisa1-formonX with coecientsr in E: r 2 1 () A (E)=(T ⇤ E)=Hom (T ,E). r 2 X ⌦R R X We interpret () as the derivative of . r Analogy:Letf : X R be a smooth function. Then D(f)=df : TX R is a 1-form on X: ! !

df TX⇤ =Hom(TX , R). 2 Given a vector field on X, D (f)=df ( )isthedirectional derivative of f in the 0 direction .Thisisateveryx X.So,D (f) ⌦X is a smooth function on X. For a connection we write: 2 2 ():= () E. r r 2 This is the -directional derivative of the section in the direction of the tangent vector field . r MATH 250B: COMPLEX ALGEBRAIC GEOMETRY 23 Following our philosophy of concentrating on concepts and definitions (and skipping proofs of theorems), we reformulate the definition of a connection.

Lemma 3.4.1. If E,E0 are vector bundles on X then

Hom (E,E0)=Hom (E,E0) = Hom 0 (E,E0). X R ⌦X k ` Proof. Locally, a homomorphism of vector bundles E E0 is given by a map ! M : X Hom (Rk, R`) ! R which is a family of ` k matrices M(x), with entries mij : X R. ⇥ ! Locally, sections of E = X Rk are given by k functions on X: ⇥ u E =(u , ,u )= u e 2 1 ··· k i i 0 X 0 where ei are “basic sections” of E and ui ⌦X .I.e.,E is a free ⌦X -module. An 0 2 ⌦ -morphism M :E E0 is therefore given by: X ! M()= u M(e )= u m e0 =(u , ,u )M i i i ij j 1 ··· k So, M is given by the sameX data: a matrixX with entries m ⌦0 . ij 2 X ⇤ 1 1 0 For a connection :E A (E),E, A (E)areboth⌦X modules. But is not r0 ! r ahomomorphismof⌦X modules: (f)=df + f (). r r However, if we have another connection 0, r 0(f)=df + f 0(). r r So, the di↵erence ( 0)(f)=f( 0) rr rr is a homomorphism of ⌦X -modules. Analogously to the proof of Lemma 3.4.1 such 1 morphisms are given by matrices M =(mij)withmij ⌦X . In local coordinates, d is a connection. So, an arbitrary2 connection is given by = d + ' or: r (f , ,f )=(df , ,df )+(f , ,f )M r 1 ··· k 1 ··· k 1 ··· k 1 where M is a k k matrix with entries in ⌦X . If X has a Riemannian⇥ metric g then recall that the Levi-Civita connection = LC r r is the unique connection on E = TM having the properties: (1) dg(,⌧ )=g( ,⌧)+g(, ⌧)( is compatible with g) (2) (⌧) (r)⌧ =[,⌧ ]usuallywrittenas:r r r r (⌧) ()=[,⌧] r r⌧ for any two vector fields ,⌧ on X. As discussed in class, Equation (1) says that, for any vector field on X, dg(,⌧ )( )=g( ,⌧ )+g(, ⌧). r r 24 MATH 250B: COMPLEX ALGEBRAIC GEOMETRY

0 Explanation: Since g(,⌧) ⌦X ,allthreetermsinEquation(1)are1-formsonX.For example, if = ⇠ ↵ where2 ⇠ are vector fields and ↵ are 1-forms on X,then r i i i i P g( ,⌧)= g(⇠i,⌧)↵i. r Applying both sides to the vector field weX get: g( ,⌧ )( )= g(⇠ ,⌧)↵ ( )= g(⇠ ↵ ( ),⌧)=g( ( ),⌧)=g( ,⌧). r i i i i r r Recall that a connectionX on a smoothX bundle E over X is a linear map :E = A0(E) A1(E) r ! satisfying Leibniz rule. When X is a complex manifold and E is a holomorphic bundle, A0(E),A1(E)arethesamesetasbeforebutwithmorestructure: 1 1,0 0,1 A (E)=(T ⇤ E)=(T ⇤ E)=A (E) A (E) X ⌦R X,C ⌦C ⌦ and A0(E)=A0,0(E) is still the space of smooth sections of E. Then, any connection : A0,0(E) A1(E)=A1,0(E) A0,1(E) r ! ⌦ has two components 1,0, 0,1.Lasttimeweshowedthat r r @ : A0,0(E) A0,1(E) E ! given in local coordinates by @ (f , ,f )=(@f , , @f )iswelldefined. U 1 ··· k 1 ··· k Proposition 3.4.2. Given a Hermitian metric h on a holomorphic bundle E, there is a unique connection on E so that r (1) dh(,⌧)=h( ,⌧)+h(, ⌧) for all ,⌧ A0,0(E)( is “compatible” with h) r r 2 r (2) 0,1 = @ r E This unique connection is called the Chern connection on E. Proof. = 1,0 + 0,1 where 0,1 = @ and 1,0 is uniquely determined by: r r r r E r (3.3) dh(,⌧)=h( 1,0,⌧)+h(, @⌧) r since h(0,1,⌧)=0andh(, 1,0⌧)=0.Inmoredetail,let 1,0 = @ + M where, in 1,0 r local coordinates, M is a k k matrix with entries in ⌦ .Wehave: ⇥ X h(u, v)= hijuivj where X @ @ h = h , . ij @z @z ✓ i j ◆ Then dh(u, v)=h(@u,v)+h(u, @v)+ dhijuivj and X h( 1,0u, v)=h(@u,v)+h(uM, v). r Let M be the solution of the equation

h(uM, v)= dhijuivj which is bilinear in u, v.Then 1,0 = @ + MXwill satisfy (3.3). r ⇤ MATH 250B: COMPLEX ALGEBRAIC GEOMETRY 25

1,0 TX iTX,R iv = iv˜ = iv + Iv

v˜ = v iIv

< <

v Iv TX,R

1,0 Figure 1. (Showing that ( )isaconnectiononTX,R.) Since iv˜ = < r 1,0 iv + Iv = Iv, (iv˜)=I (˜v)making : TX TX,R an isomorphism of < < < 1!,0 complex vector spaces with i acting as i on TX and as I on TX,R. f Theorem 3.4.3. Let (M,I) be a complex manifold with a Hermitian metric h = g i! and let LC be the Levi-Civita connection for g. Then the following are equivalent. r (1) h is a K¨ahlermetric (d! =0). (2) LC (I)=I LC () for every real vector field on X. (3) Ther holomorphicr part of the Chern connection is equal to the Levi-Civita con- nection: ( 1,0)= LC < r r (See Fig 1.) So, the Chern connection on T is “ ( LC , @) ”. X,C r Proof. (3) (2). The Chern connection is complex linear and the holomorphic part is the part where) i = I: (i)=I () 1,0 < < for asectionofTX .So, LC (I)= ( 1,0(I)) = (i 1,0()) = I ( 1,0()) = I LC (). r < r < r < r r (2) (1). Since = LC is compatible with g = (h), we are given that ) r r < d(g(,⌧)) = g( ,⌧)+g(, ⌧). r r Replacing ⌧ with I⌧ we get g(,I⌧)=!(,⌧)and I⌧ = I ⌧ by (2). So, r r d(!(,⌧ )) = !( ,⌧)+!(, ⌧). r r Apply both to vector field ,useruledf ()=(f)andskew-symmetryof!: (!(,⌧)) = !( ,⌧ ) !( ⌧,). r r Cyclically permute the three vector fields: (!(⌧,)) = !( ⌧,) !( ,⌧) r r 26 MATH 250B: COMPLEX ALGEBRAIC GEOMETRY ⌧(!(,)) = !( ,) !( ,). r⌧ r⌧ Now use (2.1), the coordinate invariant definition of d!: d!(,,⌧)=(!(,⌧)) (!(,⌧)) + ⌧(!(,)) !([,],⌧)+!([,⌧],) !([,⌧ ],) Since [,]= and !(,⌧)=!(⌧,), this is: r r d!(,,⌧)=(!(,⌧ )) + (!(⌧,)) + ⌧(!(,)) +!( () (),⌧)+!( (⌧) (),)+!( () (⌧),)=0. r r r r⌧ r⌧ r (1) (3). The proof is by reduction to the standard case: When the metric h is ) constant, LC = d and Ch =(@,@)and (@)=d.So,(3)holds.Sincethesearefirst order di↵erentialr equations,r it suces for h

✏ij0 = ✏ji The key property of these numbers is: Claim: If h is K¨ahler then k i ✏ij = ✏kj

Proof: Since @ =0and@z =0 @✏0 =0,atthepointz =0wehave: ij k ) ij i i 0=@! = @✏ dz dz = ✏k dz dz dz 2 ij i ^ j 2 ij k ^ i ^ j ij X Xijk i i where we recall that the 2 factor comes from: (z)= 2 (z z)(andd! =0isequivalent = k i to @! =0=@!). In order for these terms to cancel, we must have ✏ij = ✏kj as claimed. MATH 250B: COMPLEX ALGEBRAIC GEOMETRY 27 Now let 1 k z0 = z + ✏ z z . j j 2 ij i k k Since ✏ijzizk is symmetric in zi,zk,wehave X k dz0 = dz + ✏ z dz = dz + ✏ dz = dz + O( z ) j j ij k i j ij i j | | which implies; X X 2 dz = dz0 ✏ dz0 + O( z ) j j ij i | | @ @zj @ X @ 2 = = (ij ✏ij) + O( z ) @zi0 @zi0 @zj @zj | | So, up to terms of second order,X we have:X @ @ hij0 = h , = (ik ✏ik)hk`(j` ✏j`) @zi0 @zj0 ✓ ◆ Xk,` = ( ✏ )( + ✏ + ✏0 )( ✏0 ) ik ik k` k` k` j` `j Xk,` since ✏j` = ✏`0 j,

= ✏ + ✏ + ✏0 ✏0 ik k` j` ik k` j` ik k` j` ik k` j` ik k` j` X = ✏ + ✏ + ✏0 ✏0 = ij ij ij ij ij ij In other words, the matrix (hij0 )ofh with respect to the new coordinates zj0 is equal to the identity matrix up to second order. This proves the Lemma and completes the proof of Theorem 3.4.3. ⇤ 28 MATH 250B: COMPLEX ALGEBRAIC GEOMETRY 3.5. Examples of K¨ahler manifolds. An easy example is a Riemann surface. This is a complex 1-dimensional and real 2-dimensional manifold. Any Hermitian metric is K¨ahler since all 2-forms on a real 2-dimensional manifold are closed. The next example is CP n = Pn(C). We will construct the Fubini-Study metric on complex projective space Pn(C)andshowedthatitisaK¨ahlermetric.Thiswillimply that all smooth projective varieties over C are K¨ahlermanifolds. The outline of the construction is: L (L, h) ! h 7! 7! L $ ! Given a holomorphic line bundle L on a complex manifold X,choseahermitianformh on L.Then,thereisanassociated2-form!L on X (called the Chern form of (L, h)). This 2-form !L is associated to a Hermitian metric h! on X (h! = h)which,ifweare lucky, will be positive definite and therefore a K¨ahler metric. We6 will apply this to the n n canonical line bundle S⇤ over X = P (C)toobtaintheFubini-StudymetriconP (C). A line bundle L over X is the union over open sets Ui of Ui C.ForeachUi,take ⇥ the unit section i(v)=(v, 1). These in general don’t match. So, there are functions gij : Ui Uj C⇥ so that \ ! i(v)=gij(v)j(v) for all v U U .Since = g we have = g = g g .So, 2 i \ j j jk k i ij j ij jk k gik = gijgjk on Ui Uj Uk.Conversely,anycollectionofmapsgij : Ui Uj C⇥ satisfying the \ \ \ ! equations in the box will uniquely determine a line bundle. If the gij are holomorphic functions, the line bundle will be a holomorphic bundle. 1 For example, g⇤ := is another collection of functions satisfying the same (boxed) ij gij equations. So, gij⇤ gives another holomorphic line bundle L⇤ which one can show is the dual bundle{ to }L. + Let h be a Hermitian metric on L.Lethi : Ui R be the positive function given ! by hi(v)=h(i(v),i(v)). Since i = gijj we get:

hi = h(i,i)=gijgijh(j,j)=gijgijhj + Lemma 3.5.1. Conversely, any family of functions hi : Ui R satisfying the equa- ! tions hi = gijgijhj gives a Hermitian metric on L.

Proof. Let hi0 be another collections of functions so that hi0 = gijgijhj0 .OneachUi let fi = hi0 /hi.Thenfj = hj0 /hj = gjigjihi0 /gjigjihi = hi0 /hi = fi.So,f = fi = fj is a globally defined function on X and h0 = fh is another metric on L. ⇤ 1 For example, h⇤ = satisfies i hi

hi⇤ = gij⇤ gij⇤hj⇤

Therefore, hi⇤ gives a metric on L⇤. Let 1 ! = @@ log h . i 2⇡i i MATH 250B: COMPLEX ALGEBRAIC GEOMETRY 29 Note that

log hi =loggij +loggij +loghj.

Since gij is holomorphic, @ log gij =0.Sincegij is antiholomorphic, @ log gij =0.So, 1 1 ! = @@ log h = @@ log h = ! . i 2⇡i i 2⇡i j j 2 2 So, ! = !i is a well-defined 2-form on all of X. Also d! = @!+@! =0since@ =0=@ . Theorem 3.5.2. Given a holomorphic line bundle L on a complex manifold X and a Hermitian metric h on L, there is a closed form ! on X of type (1, 1) given locally by 1 ! = @@ log h. 2⇡i We call ! the Chern form of (L, h).

Now let X = Pn(C). Recall that this is the quotient space of Cn+1 0 modulo the relation \ (z ,z , ,z ) (z , ,z ) 0 1 ··· n ⇠ 0 ··· n for any =0 C. The equivalence class is denoted [z0, ,zn]. Another interpretation 6 2 ··· is that Pn(C)isthesetofonedimensionalsubspacesofCn+1.Eachsuchisuniquely determined by any nonzero vector (z0, ,zn) andwemaketheidentification =[z , ,z ]. ··· 2 0 ··· n Let S be the tautological line bundle over Pn(C)givenby S = (,v) Pn(C)andv Pn(C) Cn+1. { | 2 2 }⇢ ⇥ This is “tautological” since the fiber over the point Pn(C)isthespace Cn+1. n 2 ⇢ Let Ui be the open subset of P given by U = [z] z =0 . i { | i 6 } Let i be the section of S over Ui given by z z z () = ([z , ,z ]) = 0 , , i =1, , n . i i 0 ··· n z ··· z ··· z ✓ i i i ◆ This is well-defined since, e.g., the jth coordinate is z z j = j zi zi

i() is the unique element of with ith coordinate equal to 1. Comparing this with z z z ([z]) = 0 , , i , , n j z ··· z ··· z ✓ j j j ◆ we see that zj i = j. zi

So, the transition functions for S are gij = zj/zi with dual gij⇤ = zi/zj. 30 MATH 250B: COMPLEX ALGEBRAIC GEOMETRY

Since the line bundle S is a subbundle of the trivial bundle Pn Cn+1 it gets a n+1 ⇥ metric by restricting the standard metric on C given by h(z,z0)= zjzj0 .Soh(z)= h(z,z)= z 2.Sincetheith coordinate of is 1 we get: | j| i P 2 P h(i)=1+ zj . | | j=i X6 n On the dual bundle S⇤ (called the canonical bundle over P ) we have 1 h⇤(⇤)= . i 1+ z 2 | j| 2 Using z = z z ,theChernformofS⇤ on U is | j| j j i P 1 1 ! = @@ log . i 2⇡i 1+ z z ✓ j j ◆ We calculated this step by step using first the equationP 1 @f 1 @f @ log = = dzj f f f @zj to get: X 1 z dz @ log = j j . 1+ z z 1+ z 2 ✓ j j ◆ P | j| Apply @ using the quotient rule to get: P P 1 (1 + z 2) dz dz z z dz dz @@ log = | j| j ^ j i j i ^ j 1+ z z (1 + z 2)2 ✓ j j ◆ ✓ P P | j| P ◆ @f where we used the formula @(fdzj)= dzi dzj.Attheoriginz =0weget P @zi ^ P i ! = P dz dz 2⇡ j ^ j which is the standard form correspondingX to (a positive scalar multiple of) the standard metric with matrix equal to the identity matrix (divided by ⇡). So, the corresponding n metric h! is positive definite at the point z = 0. However, the space P (C)ishomoge- neous (the same at every point). This is easier to see if we use a vector space without a basis: Let V be any n +1dimensionalvectorspaceoverC and let P(V )bethespace of 1-dimensional subspaces of V .Thenitisclearthateverypointisthesameasevery other point. The tautological bundle S and its dual are also defined without choice of coordinates. So, we can choose coordinates to make any point the center point z =0. So, the canonically defined metric h! is positive definite at every point.

Theorem 3.5.3. The hermitian form h! corresponding to the canonical Chern form ! n on the dual S⇤ of the tautological line bundle over P (C) is positive definite and therefore a K¨ahlermetric. n This form h! is called the Fubini-Study metric on P (C). Corollary 3.5.4. Every smooth projective variety over C is a K¨ahlermanifold.