18.755 second problems solutions 1. Suppose V is a finite-dimensional real vector space, and φ: R → GL(V ) is a continuous homomorphism, and ǫ ∈ Cc∞ (R) is a smooth function of compact support. For v ∈ V , define vǫ = Z ∞ Z ∞ ǫ(t)φ(t)v dt. −∞ (1) Prove that φ(s)vǫ = ǫ(t − s)φ(t)v dt. −∞ In order to apply a linear transformation to an integral of a vector-valued continuous function over a compact set, you can apply the linear transformation under the integral sign. This is a generalization of A Z b f (t) dt + B a Z b g(t) dt = a Z b (Af + Bg) dt, a (from 18.01) with exactly the same proof. Since φ(s) is linear, we get ∞ φ(s)vǫ = Z = Z ǫ(t)[φ(s)φ(t)v] dt = Z ǫ(t)[φ(t + s)v] dt = Z ǫ(t′ − s)[φ(t′ )v] dt′ φ(s)[ǫ(t)φ(t)v] dt −∞ ∞ −∞ ∞ because φ is a homomorphism −∞ ∞ change of variable t′ = t + s −∞ This is the desired formula. (2) Prove that s 7→ φ(s)vǫ is a smooth map from R to V . You learn in 18.100 that a Riemann integral of two-variable function can be differentiated under the integral sign with respect to the extra variable (for example Theorem 9.42 in Rudin’s book). The conclusion is dm [φ(s)vǫ ]/dsm = Z ∞ (−1)m [dm ǫ(t − s)/dsm ][φ(t)v] dt −∞ This exhibits all the derivatives of φ(s)vǫ as continuous functions of s. (3) Prove that φ is smooth. 1 2 A matrix valued-function like φ is smooth if and only if all the vector-valued functions φv are smooth. (These are linear combinations of the columns of φ. Define V ∞ = {v ∈ V | φ(s)v is smooth}; our job is to show that V ∞ = V . The second part of the problem showed that each vector vǫ ∈ V ∞ . Suppose we choose some norm k · k on V . For any ǫ′ > 0 (sorry that I already used ǫ!) choose δ > 0 so small that kφ(t)v − vk < ǫ′ (|t| < δ); this is possible by the continuity of φ, and the fact that φ(0) is the identity. Now choose the function ǫ to be non-negative, supported on [−δ, δ], and satisfying Z ∞ ǫ(t) dt = 1. −∞ 2 (This is easy to do: you can even write a formula for ǫ using e−1/x .) Then Z ∞ kvǫ − vk = k ǫ(t)[φ(t)v − v] dtk −∞ Z ∞ ǫ(t)k[φ(t)v − v]k dt ≤ −∞ Z ∞ ≤ ǫ′ ǫ(t) dt = ǫ′ −∞ That is, we can make vǫ as close to v as we wish. This proves that the subspace V ∞ of V is dense in V . But the only dense subspace of a finite-dimensional real vector space is the whole space; so V ∞ = V , as we wished to show. 2. Suppose A is an n × n real matrix. Find necessary and sufficient conditions on A for the one-parameter group {exp(tA) | t ∈ R} to be closed in GL(n, R). Here is the answer: exp(RA) is closed if and only if A is diagonalizable as a complex matrix, and all the nonzero eigenvalues are purely imaginary numbers iyj , and all the ratios yj /yk are rational numbers. The proof requires some detailed understanding of Jordan canonical form for real matrices. I will just quote a useful version of this, without helping you find a reference for exactly this statement. Theorem. Suppose A is a linear transformation on a finite dimensional real vector space V . Then there is a unique decomposition A = Ah + Ae + An subject (1) (2) (3) to the requirements the linear transformations Ah , Ae , and An commute with each other; the linear transformation Ah is diagonalizable with real eigenvalues; the linear transformation Ae is diagonalizable over C, with purely imaginary eigenvalues; and (4) the linear transformation An is nilpotent: AN = 0 for some N > 0. 3 The subscripts h, e, and n stand for “hyperbolic,” “elliptic,” and “nilpotent.” Suppose f is a continuous map from R to a metric space. The image f (R) can fail to be closed only if there is an unbounded sequence of real numbers ti such that f (ti ) converges in the metric space. (You should think carefully about why this is true: the proof is very short, but maybe not obvious.) So if the image is not closed, then we can found an unbounded sequence ti so that exp(ti A) is convergent in GL(n, R), and in particular is a bounded sequence of matrices. By passing to a subsequence, we may assume that all ti have the same sign. Since matrix inversion is a homeomorphism, exp(−ti A) is also a (convergent and) bounded sequence of matrices. Perhaps replacing the sequence by its negative, we may assume all ti > 0. Now the Jordan decomposition guarantees exp(tA) = exp(tAh ) exp(tAe ) exp(tAn ). In appropriate coordinates the matrix Ae is block diagonal with blocks 0 yj −yj 0 (with yj 6= 0) and zeros; so k exp(tAe )k is bounded. The power series for exp(tAn ) ends after the term tN AN /N !; so k exp(tAn )k has polynomial growth in t. If Ah has a positive eigenvalue, then exp(tAh ) grows exponentially in t, so the sequence exp(ti A) cannot be bounded. Similarly, if Ah has a negative eigenvalue, then exp(−ti A) grows exponentially. The conclusion is that if the image is not closed, then Ah = 0. N +1 = 0. Then exp(tAn ) grows In exactly the same way, suppose AN n 6= 0 but An like a polynomial of degree exactly N ; so (because of the boundedness of exp(tAe )) we conclude that exp(tA) also grows like a polynomial of degree exactly N . The conclusion is that if the image is not closed, then N = 0, which means An = 0. We have shown that the image can fail to be closed only if A = Ae . In this case the image is bounded; so it is closed if and only if it is compact. Suppose that the eigenvalues of A = Ae are iyj as above, so that exp(tA) has diagonal blocks cos(tyj ) sin(tyj ) . − sin(tyj ) cos(tyj ) If all the ratios yj /y1 = pj /qj are rational, then it’s easy to see that exp(tA) is periodic with period (dividing) (least common multiple of all qj )(2π/y1 ); so the image is a circle (or a point), and is closed. Conversely, suppose that the image is compact. More or less the example done in class shows that the group cos(ty1 ) sin(ty1 ) 0 − sin(ty1 ) cos(ty1 ) cos(ty2 ) sin(ty2 ) 0 − sin(ty2 ) cos(ty2 ) is compact if and only if y1 /y2 is rational. (I’m tired of typing, so I won’t write out a proof.) By projecting the (assumed compact) {exp(tA)} on various collections of four coordinates, and using “continuous image of compact is compact,” we deduce that all the ratios yj /yk are rational, as we wished to show.

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