2 Constructions of manifolds. (Solutions)

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1 2 Constructions of manifolds. (Solutions) Last updated: February 16, Problem 1. The state of a double pendulum is entirely defined by the positions of the moving ends of the two simple pendula of which the double pendulum is constructed. It is given by two angles defined up to integral multiples of 2π. Hence the configuration space is S 1 S 1 = T 2. Problem 2. The desired diffeomorphism sends a point with polar coordinates (r, θ) of the plane without the origin to the point (e iθ, ln r) S 1 R. The inverse map is (e iθ, z) (r = e z, θ) (using polars) or (e iθ, z) (e z cos θ, e z sin θ) R 2 \ {0} using Cartesian coordinates. Problem 3. Consider the equation t 2 x 2 y 2 = ε. Different cases are distinguished by the sign of ε. If ε = α 2 > 0, then we can introduce new Cartesian coordinates by setting t = αt, x = αx, and y = αy, in which our equation is equivalent to T 2 X 2 Y 2 = 1. This is a two-sheeted hyperboloid (we omit a picture). In particular, T 2 = 1+X 2 +Y 2 1. Its vertical sections through the T -axis are hyperbolae and the horizontal sections are circles when they are non-empty or not single points (1, 0, 0), 1, 0, 0 (the two apices). Now, if ε = α 2 < 0, then in the new Cartesian coordinates the equation is equivalent to T 2 X 2 Y 2 = 1. This is a one-sheeted hyperboloid. All its horizontal sections are circles and the vertical sections through the T -axis are hyperbolae. In both cases, we have d(t 2 x 2 y 2 ε) = 2t dt 2x dx 2y dy, which vanish only at (t, x, y) = 0, which does not belong to the surface. So we have a non-degenerate equation, which specifies a closed submanifold (of dimension 2) in R 3. An exceptional case is ε = 0. Then the equation is t 2 x 2 y 2 = 0. It specifies a round cone with the axis being the t-axis. Since the point 0 belongs to this surface, the condition of non-degeneracy is violated at this point, and the cone is not a submanifold. (It becomes a submanifold if the offending point 0 is removed. The we have the disjoint union of the two half-cones without the apex.) Problem 4. Let ( ) a11 a g = 12. a 21 a 22 Consider the function f(g) = a 11 a 22 a 12 a For its differential we obtain the expression df = da 11 a 22 + da 22 a 11 da 12 a 21 da 21 a 12. We can write this in the matrix form, as ( ) ( ) f a22 a = 21. a 12 a 11 a ij 1

2 Clearly df 0 on matrices g such that det g = 1. So the equation det g = 1 is non-degenerate and therefore the subset SL(2) specified by it (which the special linear group) is a closed submanifold in GL(2). Its dimension is = 3. (An answer to the question in brackets is that differentiating the determinant of a matrix gives, up to a sign, the algebraic adjunct of an element, i.e., the minor obtained by deleting the corresponding row and column with the sign ( 1) i+j. For matrices 2 2 the minors of course reduce to single entries, as we have above. It is known that the transpose of the matrix of algebraic adjuncts is the inverse matrix multiplies by the determinant.) Problem 5. Consider, for example, a 23 To calculate it, we expand the determinant over the second row: = a 21 a 12 a 13 a 32 a 33 + a 22 a 11 a 13 a 31 a 33 a 23 a 11 a 12 a 31 a 32 (The sign of the first term is ( 1) i 1 where i is the row number, so here it is ( 1) 1 = 1, and then the signs alternate.) Hence a 23 = a 11 a 12 a 31 a 32 = a 11 a 12 ( 1)2+3 a 31 a 32 It is clear that in general, a 23 = ( 1)i 1+j 1 M ij = ( 1) i+j M ij, where M ij is the ijth minor, i.e., the determinant of the square submatrix obtained by deleting the ith row and the jth column from A. Hence the partial derivatives of det A may be written as a matrix as follows: ( ) M det A 11 M 12 M 13 = M 21 M 22 M 23. a ij M 31 M 32 M 33 We need to show that the matrix entries of this matrix cannot vanish simultaneously. This follows from the relation with the inverse matrix. If we denote 2

3 (adj A) ij = ( 1) i+j M ij and consider the matrix adj A with such entries (adj is pronounced adjunct ), then A (adj A) T = det A E or A 1 = 1 det A (adj A)T. This is a well-known theorem in algebra. Indeed, consider the matrix product a 11 a 12 a 13 M 11 M 12 T M 13 a 11 a 12 a 13 M 11 M 21 M 31 = a 31 a 32 a 33 M 21 M 22 M 23 M 31 M 32 M 33 a 31 a 32 a 33 Calculate, for example, its entry with the indices 11. We obtain a 11 M 11 a 12 M 12 + a 13 M 13 = det A. Now, the entry with the indices 12. We obtain a 11 M 21 + a 12 M 22 a 13 M 23 = = 0 M 12 M 22 M 32 M 13 M 23 M 33 (as a determinant with two repeating rows). So the ijth entry of the product is either det A when i = j (as the expansion of det A over the ith row) or zero when i j (as the expansion of the determinant obtained from det A by replacing the jth row by the ith row). So indeed A (adj A) T = det A E. Therefore the matrix adj A = (det A/a ij ) cannot be identically zero when det A = 1. So the equation det A = 1 is non-degenerate and specifies a closed submanifold in Mat(3), of dimension = 8. Obviously, the same argument works for SL(n) for arbitrary n. Problem 6. Consider the set of all orthogonal matrices in Mat(n), which is specified by the equation AA T = E. It is equivalent to A 1 = A T. (Note that if B satisfied the same equation, then AB(AB) T = ABB T A T = AEA T = AA T = E, so this subset is closed under matrix product. Now, E is obviously orthogonal, and (A T ) 1 = (A 1 ) T = (A T ) T, so the inverse of an orthogonal matrix is also orthogonal. So we indeed have a group.) The auxiliary linear equation for AA T = E is ȦA T + A A T = 0. (Here (A T ) = ( A) T, so we denote it unambiguously as A T.) Note that AȦT = ( AA T ) T. Therefore the auxiliary equation can be re-written equivalently as ( AA T ) T = ȦAT. We can establish an isomorphism between its solution space and the antisymmetric matrices as follows. Each A satisfying this equation we map 3.

4 to B = ȦAT = AA 1, where B T = B. Conversely, each B such that B T = B is mapped to A = BA. These two linear maps are mutually inverse and establish the desired isomorphism of vector spaces. Therefore the dimension of the solution space does not depend on A O(n), hence the rank of the auxiliary linear system (which is write as a single matrix equation) is constant. By the theorem about equations of constant rank, the matrix equation AA T = E specifies a closed submanifold in Mat(n). Hence the orthogonal group O(n) is a smooth manifold. Its dimension equals the dimension of the solution space of the auxiliary system, i.e., the dimension of the vector space of all antisymmetric matrices n n. It is n(n 1)/2. Indeed, any such matrix has zero diagonal entries and its entries below the diagonal are the negatives of the corresponding entries above the diagonal, for which there are no further constraints. Hence we have (n 2 n)/2 independent variables. So dim O(n) = n(n 1)/2. Problem 7. (a) To establish the desired formula, we may use the definition of derivative. Consider A + εx, where ε is a small parameter, and calculate det(a + εx). We have det(a + εx) = det ( (E + εxa 1 )A ) = det(e + εxa 1 ) det A. We need to calculate det(e +εxa 1 ) retaining terms of the first degree in ε only. Denote XA 1 = B and expand det(e + εb) over the first row. Then one can observe that the only input into the terms of first degree will be given by (1 + εb 11 ) multiplied by the corresponding minor. All other terms will be of higher degree in ε. Indeed, the minor arising as the coefficient of the entry ε b 1k, where k > 1, always contains one row proportional to ε (namely, the kth row, because the kth column containing the element 1 + ε b kk is deleted and all other elements in this row are of the form ε b kl, where l k). By induction we obtain det(e + εb) = (1 + εb 11 )(1 + εb 22 )... (1 + εb nn ) + O(ε 2 ) = Therefore 1 + ε(b 11 + b b nn ) + O(ε 2 ) = 1 + ε tr B + O(ε 2 ). det(a + εx) = det A det(e + εxa 1 ) = det A (1 + ε tr(xa 1 )) + O(ε 2 ) = hence or d dt d det A = det A tr(da A 1 ), det A + ε det A tr(xa 1 ) + O(ε 2 ), ( ) da det A = det A tr dt A 1. (b) The auxiliary linear equation for the equation det A = 1 is det A tr ( Ȧ A 1) = 0 4

5 or tr ( A A 1) = 0. We may establish an isomorphism between its solution space and the space of all trace-free matrices in Mat(n), as follows. To each matrix Ȧ satisfying the above, we assign the matrix B = Ȧ A 1, which satisfies tr B = 0. Conversely, to each B such that tr B = 0 we assign A = BA, which evidently satisfies the auxiliary equation. Therefore, the solution spaces for all A are isomorphic (being isomorphic to the solution space at A = E), hence have the same dimension. So the rank of the auxiliary equation is constant, and we are in the situation of the theorem about equations of constant rank. In our particular case (one scalar equation), constant rank is equivalent to the non-degeneracy. We conclude, by the theorem, that SL(n) as a subset of Mat(n) specified by a non-degenerate equation is a closed submanifold of dimension n 2 1, as claimed. 5