Hitchin-Exterior

4 Exterior algebra 4.1 Lines and 2-vectors The time has come now to develop some new linear algebra in order to handle the space of lines in a projective space P(V). In the projective plane we have seen that duality can deal with this but lines in higher dimensional spaces behave differently. From the point of view of linear algebra we are looking at 2-dimensional vector sub- spaces U⊂V. To motivate what we shall do, consider how in Euclidean geometry we describe a 2-dimensional subspace of R3. We could describe it through its unit normal n, which is also parallel to u×vwhere uandvare linearly independent vectors in the space andu×vis the vector cross product. The vector product has the following properties: •u×v=−v×u •(λ1u1+λ2u2)×v=λ1u1×v+λ2u2×v We shall generalize these properties to vectors in any vector space V– the difference is that the product will not be a vector in V, but will lie in another associated vector space. Definition 12 Analternating bilinear form on a vector space Vis a map B:V× V→Fsuch that •B(v, w) =−B(w, v) •B(λ1v1+λ2v2, w) =λ1B(v1, w) +λ2B(v2, w) This is the skew-symmetric version of the symmetric bilinear forms we used to define quadrics. Given a basis {v1, . . . , v n},Bis uniquely determined by the skew symmetric matrix B(vi, vj). We can add alternating forms and multiply by scalars so they form a vector space, isomorphic to the space of skew-symmetric n×nmatrices. This has dimension n(n−1)/2, spanned by the basis elements Eabfora < b where Eab ij= 0 if {a, b} /negationslash={i, j}andEab ab=−Eab ba= 1. Definition 13 The second exterior power Λ2Vof a finite-dimensional vector space is the dual space of the vector space of alternating bilinear forms on V. Elements of Λ2Vare called 2-vectors. 41 This definition is a convenience – there are other ways of defining Λ2V, and for most purposes it is only its characteristic properties which one needs rather than what its objects are. A lot of mathematics is like that – just think of the real numbers. Given this space we can now define our generalization of the cross-product, called the exterior product orwedge product of two vectors. Definition 14 Given u, v∈Vtheexterior product u∧v∈Λ2Vis the linear map to Fwhich, on an alternating bilinear form B, takes the value (u∧v)(B) =B(u, v). From this definition follows some basic properties: •(u∧v)(B) =B(u, v) =−B(v, u) =−(v∧u)(B) so that v∧u=−u∧v and in particular u∧u= 0. •((λ1u1+λ2u2)∧v)(B) =B(λ1u1+λ2u2, v) =λ1B(u1, v) +λ2B(u2, v) which implies (λ1u1+λ2u2)∧v=λ1u1∧v+λ2u2∧v. •if{v1, . . . , v n}is a basis for Vthen vi∧vjfori < j is a basis for Λ2V. This last property holds because vi∧vj(Eab) =Eab ijand in facts shows that {vi∧vj} is the dual basis to the basis {Eab}. Another important property is: Proposition 15 Letu∈Vbe a non-zero vector. Then u∧v= 0 if and only if v=λufor some scalar λ. Proof: Ifv=λu, then u∧v=u∧(λu) =λ(u∧u) = 0. Conversely, if v/negationslash=λu,uandvare linearly independent and can be extended to a basis, but then u∧vis a basis vector and so is non-zero. 2 42 It is the elements of Λ2Vof the form u∧vwhich will concern us, for suppose U⊂V is a 2-dimensional vector subspace, and {u, v}is a basis of U. Then any other basis is of the form {au+bv, cu +dv}, so, using u∧u=v∧v= 0, we get (au+bv)∧(cu+dv) = (ad−bc)u∧v and since the matrix /parenleftbigg a b c d/parenrightbigg is invertible ad−bc/negationslash= 0. It follows that the 1-dimensional subspace of Λ2Vspanned byu∧vfor a basis of Uis well-defined by Uitself and is independent of the choice of basis. To each line in P(V) we can therefore associate a point inP(Λ2V). The problem is, not every vector in Λ2Vcan be written as u∧vfor vectors u, v∈V. In general it is a linear combination of such expressions. The task, in order to describe the space of lines, is to characterize such decomposable 2-vectors. Example: Consider v1∧v2+v3∧v4in a 4-dimensional vector space V. Suppose we can write this as v1∧v2+v3∧v4= (a1v1+a2v2+a3v3+a4v4)∧(b1v1+b2v2+b3v3+b4v4). Equating the coefficient of v1∧v2gives a1b2−a2b1= 1 and so ( a1, b1) is non-zero. On the other hand the coefficients of v1∧v3andv1∧v4 give a1b3−a3b1= 0 a1b4−a4b1= 0 and since ( a1, b1)/negationslash= 0,b3a4−a3b4= 0. But the coefficient of v3∧v4gives a4b3−a3b4= 1 which is a contradiction. This 2-vector is not therefore decomposable. We shall find an easier method of seeing this by working with p-vectors and exterior products. 4.2 Higher exterior powers Definition 15 Analternating multilinear form of degree pon a vector space Vis a mapM:V×. . .×V→Fsuch that 43 •M(u1, . . . , u i, . . . , u j, . . . , u p) =−M(u1, . . . , u j, . . . , u i, . . . , u p) •M(λ1v1+λ2v2, u2, . . . , u p) =λ1M(v1, u2, . . . , u p) +λ2M(v2, u2, . . . , u p) Example: Letu1, . . . , u nbe column vectors in Rn. Then M(u1, . . . , u n) = det( u1u2. . . u n) is an alternating multilinear form of degree n. The set of all alternating multilinear forms on Vis a vector space, and Mis uniquely determined by the values M(vi1, vi2, . . . , v ip) for a basis {v1, . . . , v n}. But the alternating property allows us to change the order so long as we multiply by −1 for each transposition of variables. This means that M is uniquely determined by the values of indices for i1< i 2< . . . < i p. The number of these is the number of p-element subsets of n, i.e./parenleftbign p/parenrightbig , so this is the dimension of the space of such forms. In particular if p > n this space is zero. We define analogous constructions to those above for a pair of vectors: Definition 16 Thep-th exterior power ΛpVof a finite-dimensional vector space is the dual space of the vector space of alternating multilinear forms of degree ponV. Elements of ΛpVare called p-vectors. and Definition 17 Given u1, . . . , u p∈Vtheexterior product u1∧u2∧. . .∧up∈ΛpVis the linear map to Fwhich, on an alternating multilinear form Mtakes the value (u1∧u2∧. . .∧up)(M) =M(u1, u2, . . . , u p). The exterior product u1∧u2∧. . .∧uphas two defining properties •it is linear in each variable uiseparately •interchanging two variables changes the sign of the product 44 •if two variables are the same the exterior product vanishes. We have a useful generalization of Proposition 15: Proposition 16 The exterior product u1∧u2∧. . .∧upofpvectors ui∈Vvanishes if and only if the vectors are linearly dependent. Proof: If there exists a linear relation λ1u1+. . . λ pup= 0 with λi/negationslash= 0, then uiis a linear combination of the other vectors ui=/summationdisplay j/negationslash=iµjuj but then u1∧u2∧. . .∧up=u1∧. . .∧(/summationdisplay j/negationslash=iµjuj)∧ui+1∧. . .∧up and expand this out by linearity, each term has a repeated variable ujand so vanishes. Conversely, if u1, . . . , u pare linearly independent they can be extended to a basis and u1∧u2∧. . .∧upis a basis vector for ΛpVand is thus non-zero. 2 The exterior powers ΛpVhave natural properties with respect to linear transforma- tions: given a linear transformation T:V→W, and an alternating multilinear form MonWwe can define an induced one T∗MonVby T∗M(v1, . . . , v p) =M(Tv1, . . . , Tv p) and this defines a dual linear map ΛpT: ΛpV→ΛpW with the property that ΛpT(v1∧v2∧. . .∧vp) =Tv1∧Tv2∧. . .∧Tvp. One such map is very familiar: take p=n, so that ΛnVis one-dimensional and spanned by v1∧v2∧. . .∧vnfor a basis {v1, . . . , v n}. A linear transformation from a 45 1-dimensional vector space to itself is just multiplication by a scalar, so ΛnTis some scalar in the field. In fact it is the determinant ofT. To see this, observe that ΛnT(v1∧. . .∧vn) =Tv1∧. . .∧Tvn and the right hand side can be written using the matrix TijofTas /summationdisplay i1,...,i nTi11vi1∧. . .∧Tinnvin=/summationdisplay i1,...,i nTi11. . . T innvi1∧. . .∧vin. Each of the terms vanishes if any two of i1, . . . , i nare equal by the property of the exterior product, so we need only consider the case where ( i1, . . . , i n) is a permutation of (1, . . . , n ). Any permutation is a product of transpositions, and any transposition changes the sign of the exterior product, so ΛnT(v1∧. . .∧vn) =/summationdisplay σ∈Snsgn(σ)Tσ(1)1Tσ(2)2. . . T σ(n)nv1∧. . .∧vn which is the definition of the determinant of Tij. From our point of view the deter- minant is naturally defined for a linear transformation T:V→V, and what we just did was to see how to calculate it from the matrix of T. We now have vector spaces ΛpVof dimension/parenleftbign p/parenrightbig naturally associated to V. The space Λ1Vis by definition the dual space of the space of linear functions on V, so Λ1V=V/prime/prime∼=Vand by convention we set Λ0V=F. Given pvectors v1, . . . , v p∈V we also have a corresponding vector v1∧v2∧. . .∧vp∈ΛpVand the notation suggests that there should be a product so that we can remove the brackets: (u1∧. . .∧up)∧(v1∧. . . v q) =u1∧. . .∧up∧v1∧. . . v q and indeed there is. So suppose a∈ΛpV, b∈ΛqV, we want to define a∧b∈Λp+qV. Now for fixed vectors u1, . . . , u p∈V, M(u1, u2, . . . , u p, v1, v2, . . . , v q) is an alternating multilinear function of v1, . . . , v q, so if b=/summationdisplay j1<...<j qλj1...jqvj1∧. . .∧vjq then /summationdisplay j1<...<j qλj1...jqM(u1, . . . , u p, vj1, . . . , v jq) 46 only depends on band not on the particular way it is written in terms of a basis {v1, . . . , v n}. Similarly if a=/summationdisplay i1<...<i pµi1...ipui1∧. . .∧uip then /summationdisplay i1<...<i pµi1...ipM(ui1, . . . , u ip, v1, . . . , v q) only depends on a. We can therefore unambiguously define a∧bby its value on an alternating p+q-form Mas (a∧b)(M) =/summationdisplay i1<..<i p;ji,..<j qµi1...ipλj1...jqM(ui1, . . . , u ip, vj1, . . . , v jq). The product just involves linearity and removing the brackets. Example: Suppose a=v1+v2,b=v1∧v3−v3∧v2, with v1, v2, v3∈Vthen a∧b= (v1+v2)∧(v1∧v3−v3∧v2) =v1∧v1∧v3−v1∧v3∧v2+v2∧v1∧v3−v2∧v3∧v2 =−v1∧v3∧v2+v2∧v1∧v3 =v1∧v2∧v3−v1∧v2∧v3= 0 where we have used the basic rules that a repeated vector from Vin an exterior product gives zero, and the interchange of two vectors changes the sign. Note that u1∧u2∧. . .∧up∧v1∧. . .∧vq= (−1)pv1∧u1∧u2∧. . .∧up∧v2∧. . .∧vq because we have to interchange v1with each of the p u i’s to bring it to the front, and then repeating u1∧u2∧. . .∧up∧v1∧. . .∧vq= (−1)pqv1∧. . .∧vq∧u1∧u2∧. . .∧up. This extends by linearity to all a∈ΛpV, b∈ΛqV. We then have the basic properties of the exterior product; •a∧(b+c) =a∧b+a∧c 47 •(a∧b)∧c=a∧(b∧c) •a∧b= (−1)pqb∧aifa∈ΛpV, b∈ΛqV What we have done may seem rather formal, but it has many concrete applications. For example if a=x∧ythen a∧a=x∧y∧x∧y= 0 because x∈Vis repeated. So it is much easier to determine that a=v1∧v2+v3∧v4from the Exercise above is not decomposable: (v1∧v2+v3∧v4)∧(v1∧v2+v3∧v4) = 2v1∧v2∧v3∧v4/negationslash= 0. 4.3 Decomposable 2-vectors A line in P(V) defines a point in P(Λ2V) defined by a decomposable 2-vector a=x∧y. We need to characterize algebraically this decomposability, and the following theorem does just that: Theorem 17 Leta∈Λ2Vbe a non-zero element. Then ais decomposable if and only if a∧a= 0∈Λ4V. Proof: Ifa=x∧yfor two vectors xandythen a∧a=x∧y∧x∧y= 0 because of the repeated factor x(ory). We prove the converse by induction on the dimension of V. If dim V= 0,1 then Λ2V= 0, so the first case is dim V= 2. In this case dim Λ2V= 1 and v1∧v2is a non-zero element if v1, v2is a basis for V, so any ais decomposable. We consider the case dim V= 3 separately now. Given a non-zero a∈Λ2V, define A:V→Λ3Vby A(v) =a∧v. Since dim Λ3V= 1, dim ker A≥2, so let u1, u2be linearly independent vectors in the kernel and extend to a basis u1, u2, u3ofV. We can then write a=λ1u2∧u3+λ2u3∧u1+λ3u1∧u2. 48 Now by definition 0 = a∧u1=λ1u2∧u3∧u1soλ1= 0 and similarly 0 = a∧u2 implies λ2= 0. It follows that a=λ3u1∧u2, which is decomposable. Now assume inductively that the theorem is true for dim V≤n−1 and consider the case dim V=n. Using a basis v1, . . . , v n, write a=n/summationdisplay 1≤i<jaijvi∧vj = (n−1/summationdisplay i=1ainvi)∧vn+n−1/summationdisplay 1≤i<jaijvi∧vj =u∧vn+a/prime where u∈Uanda/prime∈Λ2UandUis the ( n−1)-dimensional space spanned by v1, . . . , v n−1. Now 0 =a∧a= (u∧vn+a/prime)∧(u∧vn+a/prime) = 2u∧a/prime∧vn+a/prime∧a/prime. Butvndoesn’t appear in the expansion of u∧a/primeora/prime∧a/primeso we separately obtain u∧a/prime= 0, a/prime∧a/prime= 0. By induction a/prime∧a/prime= 0 implies a/prime=u1∧u2and so the first equation reads u∧u1∧u2= 0 which from Proposition 16 says that there is a linear relation λu+µ1u1+µ2u2= 0. Ifλ= 0, then u1andu2are linearly dependent so a/prime=u1∧u2= 0. This means that u=u∧vnand is therefore decomposable. If λ/negationslash= 0,u=λ1u1+λ2u2, so a=λ1u1∧vn+λ2u2∧vn+u1∧u2 and this is the 3-dimensional case which is always decomposable as we showed above. We conclude that a, in each case, is decomposable. 2 49 4.4 The Klein quadric The first case where we can apply Theorem 17 is when dim V= 4, to describe the projective lines in the 3-dimensional space P(V). In this case dim Λ4V= 1 with a basis vector v0∧v1∧v2∧v3ifVis given the basis v0, . . . , v 3. Fora∈Λ2Vwe write a=λ1v0∧v1+λ2v0∧v2+λ3v0∧v3+µ1v2∧v3+µ2v3∧v1+µ3v1∧v2 and then a∧a=B(a, a)v0∧v1∧v2∧v3where B(a, a) = 2( λ1µ1+λ2µ2+λ3µ3) (8) This is a non-degenerate quadratic form, and so B(a, a) = 0 defines a nonsingular quadric Q⊂P(Λ2V). Moreover, any other choice of basis rescales Bby a non-zero constant and so Qis well defined in projective space. We see then that a line /lscript⊂P(V) defines a decomposable 2-vector a=x∧y, unique up to a scalar and since a∧a= 0, it defines a point L∈Q⊂P(Λ2V). Conversely, Theorem 17 tells us that every point in Qis represented by a decomposable 2-vector. Hence Proposition 18 There is a one-to-one correspondence /lscript↔Lbetween lines /lscriptin a3-dimensional projective space P(V)and points Lin the 4-dimensional quadric Q⊂P(Λ2V). It was Felix Klein (1849–1925), building on the work of his supervisor Julius Pl¨ ucker, who first described this in detail and Qis usually called the Klein quadric . The equation of the quadric in the form (8) shows that there are linear subspaces inside it of maximal dimension 2 whatever the field. The linear subspaces all relate to intersection properties of lines in P(V). For example: Proposition 19 Two lines /lscript1, /lscript2⊂P(V)intersect if and only if the line joining the two corresponding points L1, L2∈Qlies entirely in Q. Proof: LetU1, U2⊂Vbe the two-dimensional subspaces of Vdefined by /lscript1, /lscript2. Suppose the lines intersect in X, with representative vector u∈V. Then extend to bases{u, u 1}forU1and{u, u 2}forU2. The line in P(Λ2V) joining L1andL2is then P(W) where Wis spanned by u∧u1andu∧u2. 50 Any 2-vector in Wis thus of the form λ1u∧u1+λ2u∧u2=u∧(λ1u1+λ2u2) which is decomposable and so represents a point in Q. Conversely, if the lines do not intersect, U1∩U2={0}soV=U1⊕U2. In this case choose bases {u1, v1}ofU1and{u2, v2}ofU2. Then {u1, v1, u2, v2}is a basis of V and in particular u1∧v1∧u2∧v2/negationslash= 0. A point on the line joining L1, L2is now represented by a=λ1u1∧v1+λ2u2∧v2so that a∧a= 2λ1λ2u1∧v1∧u2∧v2 which vanishes only if λ1orλ2are zero. Thus the line only meets Qin the points L1 andL2. 2 Now fix a point X∈P(V) and look at the set of lines passing through this point: Proposition 20 The set of lines /lscript⊂P(V)passing through a fixed point X∈P(V) corresponds to the set of points L∈Qwhich lie in a fixed plane contained in Q. Proof: Letxbe a representative vector for X. The line P(U) passes through Xif and only if x∈U, soP(U) is represented in the Klein quadric by a 2-vector of the form x∧u. Extend xto a basis {x, v 1, v2, v3}ofV, then any decomposable 2-vector of the form x∧ycan be written as x∧(µx+λ1v1+λ2v2+λ3v3) =λ1x∧v1+λ2x∧v2+λ3x∧v3. Thus any line passing through Xis represented by a 2-vector in the 3-dimensional space of decomposables spanned by x∧v1, x∧v2, x∧v3, which is a projective plane inQ. Conversely any point in this plane defines a line in P(V) through X. 2 A plane in Qdefined by a point X∈P(V) like this is called an α-plane . There are other planes in Q: Proposition 21 LetP(W)⊂P(V)be a plane. The set of lines /lscript⊂P(W)corre- sponds to the set of points L∈Qwhich lie in a fixed plane contained in Q. 51 A plane of this type contained in Qis called a β-plane . Proof: We just use duality here: if U⊂Vis 2-dimensional, then its annihilator U0⊂V/primeis 4−2 = 2-dimensional, so there is a one-to-one correspondence between lines in P(V) and lines in P(V/prime). A point in Qtherefore defines a line in either the projective space or its dual. Now the dual of the set of lines passing through a point is the set of lines lying in a (hyper)-plane. So applying Proposition 20 to P(V/prime) gives the result. 2 In fact there are no more planes: Proposition 22 Any plane in the Klein quadric Qis either an α-plane or a β-plane. Proof: Take a plane in Qand three non-collinear points L1, L2, L3on it. We get three lines /lscript1, /lscript2, /lscript3inP(V). Since the line joining L1toL2lies in the plane and hence in Q, it follows from Proposition 19 that each pair of /lscript1, /lscript2, /lscript3intersect. There are two possibilities: •the three lines are concurrent: •the three lines meet in three distinct points: In the first case the three lines pass through a single point and so L1, L2, L3lie in an α-plane. But this must be the original plane since the three representative vectors forL1, L2, L3are linearly independent as the points are not collinear. 52 In the second case, if u1, u2, u3are representative vectors for the three points of intersection of /lscript1, /lscript2, /lscript3, then L1, L2, L3are represented by u2∧u3, u3∧u1, u1∧u2. A general point on the plane is then given by λ1u2∧u3+λ2u3∧u1+λ1u3∧u1 which is a general element of Λ2Uwhere Uis spanned by u1, u2, u3. Thus /lscript1, /lscript2, /lscript3all lie in the plane P(U)⊂P(V). 2 The existence of these two families of linear subspaces of maximal dimension is char- acteristic of even-dimensional quadrics – it is the generalization of the two families of lines we saw on the “cooling tower” quadric surface. In the case of the Klein quadric, two different α-planes intersect in a point, since there is a unique line joining two points. Similarly (and by duality) two βplanes meet in a point. An α-plane and a β plane in general have empty intersection – if Xis a point and πa plane with X/negationslash∈π, there is no line in πwhich passes through X. IfX∈π, then the intersection is a line. 4.5 Exercises 1. Ifa∈ΛpVandpis odd, show that a∧a= 0. 2. Calculate a∧bin the following cases: •a=b=v1∧v2+v2∧v3+v3∧v1 •a=v1∧v2+v3∧v1, b =v2∧v3∧v4 •a=v1+v2+v3, b =v1∧v2+v2∧v3+v3∧v1. [v1, v2, v3, v4are linearly independent ] 3. Which of the following 2-vectors is decomposable? •v1∧v2+v2∧v3 •v1∧v2+v2∧v3+v3∧v4 •v1∧v2+v2∧v3+v3∧v4+v4∧v1. 53 [v1, v2, v3, v4are linearly independent ] 4. If dim V=nshown that every a∈Λn−1Vis decomposable. 5. Let /lscript⊂P(V) be a line and manother such that the corresponding point M∈Q lies on the polar hyperplane to L∈Q. Show that /lscriptandmintersect. 54