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Geometry of Complex Vector Spaces Stereographic

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Geometry of Complex Vector Spaces Stereographic Geometry of Complex Vector Spaces Stereographic projection. n+1 Let the coordinates in R be u0, ..., un.Thelocus u2 ··· u2 (1) 0 + + n =1 n of unit length vectors is called an n-dimensional sphere, and is often denoted by S . Its dimension n is the 2 1 number of degrees of freedom of a point on the locus. So S is the usual unit sphere in 3-space, and S is the unit circle in the plane. 3 Stereographic projection is useful for visualizing the sphere, especially the 3-sphere S . Via stereographic projection, the points of the n-sphere correspond bijectively to points of the n-dimensional hyperplane H defined by the equation u0 = 0. Though it is not traditional, I like to depict the first coordinate u0 as the “vertical” axis. The north poleis the point (1, 0,... ,0) at the top of the sphere. The stereographic n projection of a point p =(u0, ..., un)on S is the intersection of the line through the north pole and the point p with H. This projection is bijective except at the north pole, where it is not defined. One says that the north pole is sent to “infinity”. In parametric form, the line of projection is (t(u0 − 1) + 1,tu1, ..., tun), and the intersection with H is the point u1 un (2) (0,y1, ..., yn)= (0, , ..., ). 1 − u0 1 − u0 2 2 2 Writing r = y1 + ···+ yn, the inverse function sends the point (0,y1, ..., yn)to 2 r − 1 2y1 2yn (3) (u0, ..., un)=( , , ..., ). r2 +1 r2 +1 r2 +1 The complex vector spaceCn . n t Let V denote the complex vector space C . We may separate a complex vector X =(x1, ..., xn) into its real t t and imaginary parts, writing xν = aν + bν i so that X = A + Bi,whereA =(a1, ..., an) and B =(b1, ..., bn) . In this way, the complex n-dimensional vector X corresponds to a pair of n-dimensional real vectors, or to a single real vector of dimension 2n. In the long run it is better not to introduce a separate real vector, but let’s do so for now, and denote this 2n-dimensional real vector by X.Howthe entries a ν ,bν of this 2n-dimensional t real vector are arranged is arbitrary. We’ll use the arrangement X =(a1,b1,a2,b2, ..., an,bn) . n 2n Thus there is a natural bijective correspondence between C and R . This bijection is not called an n 2n isomorphism of vector spaces because the field of scalars in C is the field of complex numbers, while in R it is the field of real numbers. 1 2 Multiplication by i on Cn corresponds to a linear operator on R2n whose matrix J is a block diagonal matrix made up of 2 × 2blocksbelow. Note that J 2 = −I: � � 0 −1 (4) . 1 0 Hermitian geometry. By geometry of the complex numbers one means the geometry of the complex plane, a real two-dimensional n space. Similarly when speaking of the geometry of the n-dimensional complex space C ,wemean the geometry of the corresponding real 2n-dimensional space. So the lengthof a complex vector X is defined to 2 2 2 2 be the length of the associated real vector, the square root of a1+ b 1 + ···+ an+ b n. Using the star notation ∗ t X =(x1, ..., xn) , the length is given by the formula 2 ∗ (5) |X| = XX = x1x1 + ···+ xnxn. n This formula for length is the basis for the definition of the standard hermitian producton V = C : ∗ (6) X, Y = X Y = x1y1 + ···+ xnyn. It has the following properties: Linearityin the second variable: X, Σcν Yν =Σcν X, Yν . Conjugate Linearityin the first variable: Σcν Xν ,Y =Σcν Xν ,Y . Hermitian symmetry: Y, X = X, Y . Positivity: X, X, the square length of X,is a positive real number if X =0. To express the real and imaginary parts of the hermitian product in terms of real vectors, we write xν = t aν +bν i as before, and yν = cν +dν i,sothat Y =(c1,d1, ..., cn,dn) .Since( a−bi)(c+di)= (ac+bd)+(ad−bc)i, one sees that the real part of X, Y is (7) (a1c1 + b1d1)+··· +(ancn + bndn), which is the dot product (X · Y) of the real vectors. The imaginary part (8) (a1d1 − b1c1)+··· +(andn − bncn) can also be expressed in terms of the real vectors corresponding to X and Y . It is obtained by evaluating the skew symmetricbilinear form (9) [X , Y]=− XtJY, where J is the block diagonal matrix mentioned above. The skew symmetry relation is [Y , X]= −[X , Y]. Thus (10) X, Y =(X · Y)+[X , Y]i. Recapitulating, the real part of the hermitian product is the dot product of the corresponding real vectors, and the imaginary part is obtained from the real vectors by evaluating a certain skew symmetric form. By orthogonalityof vectors in Cn, we usually mean orthogonality with respect to the standard hermitian product: X and Y are orthogonal if X, Y =0. If X is orthogonal to Y ,thenX and Y are orthogonal real 3 vectors, because the real part of X, Y is the dot product (X · Y). This is not the only requirement, because for X, Y to vanish, the imaginary part [X , Y] must be zero too. We want to stop using special symbols for the real vectors, and from now on we’ll use the same symbol X to denote the complex vector and its associated real vector. Formula (10) becomes (11) X, Y =(X · Y )+[X, Y ]i. We just have to remember that the two products which appear on the right side of this equation are real numbers. As an exercise to accustom yourself to this notation, check the formula (12) [X, Y ]=− (X · iY ). Most important: Orthogonality has acquired two meanings. Orthogonality as complex vectors (complex orthogonality)meansthat X, Y = 0, while orthogonality as real vectors (real orthogonality)means( X ·Y )= 0. Complex orthogonality is the stronger condition. n We will denote the one-dimensional subspace of C spanned by a nonzero vector X by SpanC(X), adding a subscript C to avoid confusion. This subspace consists of all complex multiples αX,ofX .It’selements correspond with bijectively to points in the complex plane. Proposition 14.Two vectorsX, Y are complex orthogonal if and only if every pair of vectors taken from the one-dimensional complex subspacesSpanC(X) andSpanC(Y ) are real orthogonal. Proof.If X and Y are complex orthogonal, then X, Y =0. Then αX, βY = αβX, Y =0for all complex numbers α, β. All pairs of vectors in the two subspaces are complex orthogonal, and therefore real orthogonal. Conversely, suppose that all vectors in SpanC(X) are real orthogonal to all vectors in SpanC(Y ). Then (X · Y ) = 0, and also (X · iY )= −[X, Y ]= 0. So X, Y =(X · Y )+[X, Y ]i = 0, which shows that X, Y are orthogonal. Geometry ofR2 . Before describing the geometry of the complex two-dimensional vector space, we’ll warm up by reviewing 2 the geometry of the real vector space R . 2 Since R has dimension 2, its proper subspaces have dimension 1. They are the lines through the origin. t Any nonzero real vector X =(x1,x2) , spans a one-dimensional subspace, and the elements of SpanR(X) are the real multiples of X. 2 Distinct 1-dimensional subspaces meet only at the origin, so R is partitioned into its one dimensional subspaces, if we agree to leave the zero vector aside. A one-dimensional subspace can be described by its slope. If W = SpanR(X), then the slope of W is λ = x2/x1. The slope can take any value, including infinity. The special value ∞ for a slope is an unpleasant artifact of our choice of coordinate system. Using stereo- 2 1 graphic projection, the slopes, and hence the subspaces of R , correspond to points of a circle S ,whichis much nicer. With coordinates u0,u1 for the circle, the correspondence is given by formulas (1),(2): u1 (14) λ = . 1 − u0 4 and � � |λ|2 − λ u ,u 1, 2 . (15) ( 1 2)= 2 2 |λ | +1 |λ | +1 The slope λ = ∞ corresponds to the north pole (1, 0). (The absolute value signs in this formula are superfluous. I’ve put them there so that the same formula will work when λ is complex.) 2 Every point of R except the origin has a slope λ, which, by the inverse of stereographic projection, cor- 1 2 σ 1 responds to a point on the unit circle S . Sothereis amapR −→ S , defined except at the origin. Its equation is obtained by substituting λ = x2/x1 into equation (15): � � x2x2 − x1x1 2x1x2 (16) σ(X)= , . x2x2 + x1x1 x2x2 + x1x1 (Again I’ve written this formula so that it works when λ is complex.) 2 2 Restricting σ to the unit circle x1 + x2 =1 in the x-plane, we set x1 =cosθ , x2 =sinθ .Then (17) σ(X)= (cos2 θ, sin 2θ). The map slope map σ wraps the unit circle in the x-plane twice around the unit circle in the u-plane. Geometry ofC2 . 2 A modification of the above discussion carries over to the complex case. Since V = C is a two-dimensional complex vector space, every proper (complex) subspace W has dimension 1, and consists of the complex multiples of any of its nonzero vectors X. Two 1-dimensional subspaces can meet only at the origin unless they are equal. So if we ignore the zero vector, then V is partitioned into one dimensional subspaces, each of which is a complex plane SpanC(X).
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