Worksheet 5 Differential Geometry II, SS 2008

Worksheet 5
Differential Geometry II, SS 2008
K. Polthier and M. Wardetzky
Due June 3
This worksheet deals with some (more or less) elementary properties of Lie groups.
Exercise 1 (3 points)
Let X ∈ gl(n, R). Using the definition of the exponential map given in the lecture, show that
exp(X) = I + X + X2 Xn
+ · · · + + . . . .
2 n!
You have to show that exp(tX), t ∈ R, is an integral curve of X that passes through the
identity at t = 0. You should also include a short argument why the right hand side converges.
Exercise 2 (3 points)
Let X, Y ∈ gl(n, R). Show that [X, Y ] = 0 implies exp(X + Y ) = exp(X) · exp(Y ). Give an
example that in general, though, exp(X + Y ) = exp(X) · exp(Y ).
Exercise 3 (2 points)
Let X ∈ gl(n, C). Show that det(exp(X)) = e traceX . Hint: use the Jordan normal form of X.
Exercise 4 (3 points)
Let X ∈gl(n, R). Show that exp(X) has positive determinant. Furthermore, show that
−2 0
is not exp(X) for any X ∈ gl(2, R). Hint: for the latter part, use that exp(tX) is
0 −1
a one-parameter subgroup, i.e., that exp((t + s)X) = exp(tX) · exp(sX).
Exercise 5 (3 points)
A metric tensor (·, ·) on a Lie group G is called left-invariant (resp. right-invariant) if (X, Y )g =
(dLg−1X, dLg−1Y )e (resp. (X, Y )g = (dRg−1X, dRg−1Y )e) for all g ∈ G. It is called biinvariant
if it is both left- and right-invariant. The Heisenberg group, H, consists of upper
triangular 3 × 3 matrices of the form
⎛
1 x
⎝0 1
⎞
y
z⎠
, x, y, z ∈ R,
0 0 1
with the group law given by matrix multiplication. Show that H does not admit a bi-invariant
metric.

Exercise 6 (6 points)
Assume that G is connected and that the metric tensor is left-invariant. Show that the following
statements are equivalent:
(1) (·, ·) is bi-invariant.
(2) Adg acts by isometries on TeG ∼ = g for all g ∈ G.
(3) adX is skew-adjoint, i.e., (adXY, Z)e + (Y, adXZ)e = 0 for all X, Y, Z ∈ g.
(4) The covariant derivative induced by (·, ·) satisfies 2∇XY = [X, Y ] for all X, Y ∈ g.
(5) The curves exp(tX) are geodesics for all X ∈ g.
Hints: (1) =⇒ (2) is rather straightforward. To show (2) =⇒ (3), use that Ad exp(tX) is an
isometry and differentiate with respect to t. To obtain (3) =⇒ (4), use the Koszul formula.
To show (4) =⇒ (5), notice that ∇XX = 0. Vice-versa, use polarization to deduce that
(5) =⇒ (4), and use the Leibnitz rule to show (4) =⇒ (3). To obtain (3) =⇒ (2), differentiate
(Ad exp(tX)(Y ), Ad exp(tX)(Z)) to show that Adg is an isometry for all g in a neighborhood U
of e and use that any element of G can be expressed as a finite product of elements of U.
Finally, (2) =⇒ (1) is rather straightforward again.