Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Singular-value decomposition wikipedia , lookup
Jordan normal form wikipedia , lookup
Capelli's identity wikipedia , lookup
Cartesian tensor wikipedia , lookup
Bra–ket notation wikipedia , lookup
Matrix calculus wikipedia , lookup
Four-vector wikipedia , lookup
Gaussian elimination wikipedia , lookup
Cayley–Hamilton theorem wikipedia , lookup
Linear algebra wikipedia , lookup
Orthogonal matrix wikipedia , lookup
We consider the projection ⇡ : SO(3) ! S 2 , where we send a matrix to its first row vector. Since matrices in SO(3) are such that all the row and column vectors are of unit length, we see that this map is well-defined. (a) We will show that ⇡ is a principal S 1 bundle. For this, we first need to define the action of S 1 on SO(3). We define it as follows: 0 1 0 1 a a @ v A · (z1 + iz2 ) := @ z1 v + z2 w A , w z2 v + z1 w where we denote with a, v, w row-vectors and z1 + iz2 2 S 1 ⇢ C an arbitrary element. Note that using that we can characterize z1 + iz2 = ei✓ , we find that ! ✓ ◆ ✓ ◆ a a cos(✓) sin(✓) i✓ ^ i✓ = ·e = ^ , with A · e Ã. sin(✓) cos(✓) à A · ei✓ Note that we have the following computations: 00 1 1 0 1 a a det @@ v A · (z1 + iz2 )A = det @ z1 v + z2 w A = z12 det(A) + z22 det(A) = det(A) = 1, w z 2 v + z1 w with some standard rule of taking the determinant, ha, z1 v + z2 wi = z1 ha, vi + z2 ha, wi = 0 + 0 = 0, ha, z2 v + z1 wi = z2 ha, vi + z1 ha, wi = 0 + 0 = 0, hz1 v + z2 w, z1 v + z2 wi = z12 hv, vi + z22 hw, wi + 0 = 1, h z2 v + z1 w, z2 v + z1 wi = z22 hv, vi + z12 hw, wi + 0 = 1. These computation show that A · (z1 + iz2 ) 2 SO(3). Let us check that this is indeed an action on SO(3). Firstly, if ei✓ = 1, then we matrix multiply with the identity matrix so that A · 1 = A for all A. Secondly, if we have g, h 2 S 1 , represented by ✓g , ✓h , then: 0 1 a ✓ ◆ A·h (A · g) · h = @ cos(✓g ) sin(✓g ) à sin(✓g ) cos(✓g ) 0 1 a ✓ ◆✓ ◆ A cos(✓g ) sin(✓g ) = @ cos(✓h ) sin(✓h ) à sin(✓h ) cos(✓h ) sin(✓g ) cos(✓g ) 0 1 a ✓ ◆ A = @ cos(✓h + ✓g ) sin(✓h + ✓g ) à sin(✓h + ✓g ) cos(✓h + ✓g ) = A · (g · h), where we use goniemetric identities and that S 1 is additive with respect to the ✓. So indeed, this is an action. Now let us check whether this action makes SO(3) into a principal bundle over S 2 . First of all, ⇡ is surjective. Given an element a 2 S 2 ⇢ R3 , we can use Gramm-Schmidt to get an orthonormal 0 1 a basis {a, v, w} of R3 such that a is the first basis elements. Then, we find that either @ v A or w 0 1 a @ v A has determinant 1, so that at least one of these two are elements of SO(3). Note that ⇡ w 1 maps them exactly to a. Secondly, we find that ⇡ is S 1 invariant by the following computation: 0 1 ✓ ◆ a ✓ ◆ A = a = ⇡ a = ⇡(A) ⇡(A · g) = ⇡ @ cos(✓) sin(✓) à à sin(✓) cos(✓) Finally, we will show that locally we can construct a section of ⇡, which is equivalent with the local triviality statement. First we let U = S 2 \ {±e3 }. If a 2 U , we know that a3 6= ±1. Hence we can define ṽa := e3 a3 a and va := |ṽṽaa | using Gramm-Schmidt. Note that since a3 6= ±1, we find that 0 1 a0 va depends smoothly on a. Now, for any a0 2 U , we pick the unique w0 such that @va0 A 2 SO(3). w0 We do this by construction, pick any element w 2 / span(a0 , av00 ) and1use Gramm-schmidt to make a0 an orthonormal basis {a0 , va0 , w0 }. Then the determinant of @va0 A is either plus or minus one, w0 so after switching w0 with w0 when the determinant was negative, gives the wanted w0 . a Now, we define w̃a := w0 hw0 , aia hw0 , va iva and wa := |w̃ w̃a | . Let us see whether we can actually do this. We see that w̃a = 0 if and only if w0 = hw0 , aia + hw0 , va iva . Since this is not the case for a = a0 and the innerproduct and va all depend smoothly on a, we see that there exists some small neighborhood V around a0 such that also wa is well-defined and hence smoothly depending on a. Now note that we can choose V to be connected, so that by continuity of the determinant and since it has only values ±1 on O(3): 0 1 0 1 a a0 det @ va A = det @va0 A = 1. wa w0 0 1 a We conclude that on V , we have the section which sends a 7! @ va A. Of course we can do the wa 2 same for all points inside S \ {±e2 }, so we conclude that around all points there exists a small neighbourhood and a local section on it. We conclude that SO(r) ! S 2 is a principal S 1 -bundle. (b) Here we will prove that ✓ E := E(SO(3), R◆2 ) is isomorphic to T S 2 . We use the reprecos(✓) sin(✓) sentation ⇢ : S 1 ! GL2 ; g 7! . Note that ⇢(g) 1 = ⇢(g)T , what we will use sin(✓) cos(✓) later on. Consider the following isomorphism: : E ! T S 2 ; [A, ⇠] 7! ⇠ T Ã. Here we view ⇠ T as a row vector and we use normal matrix multiplication. Let us first check whether everything is well-defined. So, we have to check whether [A · g, ⇢(g) 1 ⇠] gives the same answer. Here we find that ([A · g, ⇢(g) 1 ⇠]) = (⇢(g) T 1 ⇠)T · Ag · g) = ⇠ T · (⇢(g) = ⇠ ⇢(g)⇢(g) 1 T à = ⇠ à = 1 T ) ⇢(g) 1 à ([A, ⇠]). 2 So indeed, it is well-defined on the quotient. Now let us check whether the image ✓ ◆ ✓ ◆is in T S . For a v this to hold, we need that ([A, ⇠]) is orthogonal to a, with A = and à = . We compute w à that: ha, ([A, ⇠])i = ha, ⇠ T Ãi = ha, ⇠1 v + ⇠2 wi = ⇠1 ha, vi + ⇠2 ha, wi = ⇠1 0 + ⇠2 0 = 0. 2 Here we use that A 2 SO(3), so that the row columns form an orthonormal basis. We need to check a couple more things to conclude that is an isomorphism. Note that it is clear that is smooth, since it is just polynomial. Secondly, it is linear, since: ([A, ⇠] + [A, ⌘]) = ([A, ⇠ + ⌘]) = (⇠ + ⌘)T à = ⇠ T à + ⌘ T à = ([A, ⇠]) + ([A, ⌘]). And by the construction of E and the independence of choice of representative of the quotient class, this is all we need to check. We also find that this is indeed a vectorbundle morphism, that is, ⇡ = ⇡ by construction. So we are left to check whether all a are isomorphisms as linear maps. We start with injectivity. Suppose that ([A, ⇠]) = 0, this means that ⇠ T à = 0 2 R3 . Now suppose without loss of generality that ⇠1 6= 0. Then we find that since 0 = ⇠1 v + ⇠2 w, that v = ⇠⇠21 w. Note that immediately implies that ⇠2 6= 0, or alse v = 0 which is not possible since it has to have length 1. Now we find that: 0 = hv, wi = ⇠2 hw, wi = ⇠1 ⇠2 6= 0. ⇠1 Hence we find a contradiction, so that ⇠ = 0 and hence [A, ⇠] = 0. x Now let us prove surjectivity. Let x = (x1 , x2 , x3 ) 2 Ta S 2 . That is, ha, xi = 0. Now let v = |x| and use Gramm-Schmidt once more to find an w such that {a, w} is an orthonormal basis. 0v, 1 a Once more, we might have to change w into w, so that A = @ v A 2 SO(3). Now finally, let w ⇠ = (|x|, 0). Then: ([A, ⇠]) = ⇠ T à = |x| · We conclude that x + 0 = x. |x| is an isomorphism. (c) Proposition 4.43 tells us that SO(3) is isomorphic to a SO(2) structure on S 2 . Now let us consider what this exactly means. We know that a O(n) structure is the same as a metric, and that a GL+ (n) structure is the same as an orientation. Since O(n) \ GL+ (n) = SO(n), we find that finding a SO(n)-structure is the same as picking a metric and an orientation at the same time. Here use that in the linear case, a SO(n) structure is a choice of basises of V which are orthogonal and change of basis between two of them has positive determinant, so that the orthogonal basises induce an orientation. We conclude that S 2 is a oriented riemannian manifold. 3