This is for HW 6, Eerily Hot.

Question 1: Jacobi's Formula [Done need to type up]

Answer: Part A: First observe \( \textrm{det}(I_N +tA) = 1 +t \textrm{tr}(A) + \textrm{(Higher order t terms)}\)

Then differentiating with respect to \(t\) at \(t=0\)

\( \frac{d}{dt}|_{t=0} \textrm{det}(I_N +tA) = 0 +\textrm{tr}(A) + \frac{d}{dt}|_{t=0}\textrm{(Higherorder t terms)}\)

The higher order \(t\) terms vanish here since we are evaluating at \(t=0\). Leaving us with

\(\frac{d}{dt}|_{t=0}\textrm{det}(I_n+tA) = \textrm{tr}(A)\)

Part B: Observe that

\(\begin{split} \frac{d}{dt}|_{t=0}[\textrm{det}(X+tB)]&=\frac{d}{dt}|_{t=1}[\textrm{det}(X)\textrm{det}(I_n+tX^{-1}B)]\\ & =\frac{d}{dt}|_{t=0}\textrm{det}(I_n+tX^{-1}B) \times \textrm{det}(X) + 0 \\ &= \textrm{tr}(X^{-1}B)\times \textrm{det}(X) \end{split} \)

Question 2: Hopf Fibrations

Answer: We first show smoothness, which is equivalent to showing that the circle is bijective.

Observe that the group action \(\Theta: S^1 \times S^{2n-1} \rightarrow S^{2n-1}\) is given by \(\Theta_z(w) = zw\), so \(\Theta_{z^{-1}}(zw) = z^{-1}zw= (z^{-1}zw^i) = w\)

Where this multiplication is smooth on each coordinate \(w^i\)

For \(w \in S^{2n-1}\), the orbit \(S^1 \cdot w = \{z\cdot w | z\in S^1\}\) should rotate each coordinate by a whole circle

If we then look at the \((2n+1)\) sphere as a subset of \(\mathbb{C}^{n+1}\), we see that the map

\(F:S^1 (\subset \mathbb{C}) \rightarrow \mathbb{C}^{n+1} \) by \(F(z) = (zw^1, \ldots, zw^{n+1})\)

This is a linear map and is injective, so then the image is exactly a circle.

Injectivity then tells us that this map partitions the sphere into disjoint circles, so then the union of all of these circles should be \(S^{2n+1}\)

Question 3: Quaternionic Sphere

Answer:

Part A: Observe that \(p*p = (\bar{a},-b)(a,b) = (\bar{a}a +b\bar{b}, ab-ab) = (|a|^2 +|b|^2, 0)\)

So \(S \subset \mathbb{H}*\), where \(S = \{ p \in \mathbb{H} | |p| =1 \}\)

So \(S\) is a subgroup of \(\mathbb{H}*\) since for any \(p,q \in S\), we get that \(|pq| = |p||q| =1\)

The fact that \(S\) has manifold structure follows from Lee 20.12

Now we want to show that \(S \cong SU(2)\)

The isomorphism from \(SU(2)\) to \(S\) follows from the determinant map. Where \(det(A) = |p|^2 + |q|^2= aa*\)

Where \(a=(p,q) \in \mathbb{H}*\)

Part B: Let \(q=(a,b)\in S\), then \(qi = (ai,bi), qj=(-\bar{b},\bar{a}) , qk =(\bar{b}i,-\bar{a}i)\).

Taking the dot product with each other gives 0, hence this is linearly independent and all three together form a frame for \(S\). Left invariance follows from the fact that adding a left shift corresponds to a left shift in each coordinate in the frame. i.e. \(L_x(qi) = (q+x)i = qi +xi \) which is a frame for \(q+x\)

Question 4: Abelian Exponentials

Answer: Here I am referring to Lee 8.40

We have that the Lie Algebra for \(\mathbb{R}^n\) is itself and for the torus it is also \(\mathbb{R}^n\). The exponential map \(exp: \mathbb{R}^n \rightarrow \mathbb{R}^n\) is the integral curve of a constant vector field pointing in the direction of a point \(x\in \mathbb{R}^n\)

In the case of the Torus, we can think of it as \(\mathbb{R}^n\)quotiented by the basis vector in each direction with appropriate gluings. So we can then just interpret the exponential map of the torus as the quotient of \(\mathbb{R}^n\) to the torus after applying the exponential map at the origin in \(\mathbb{R}^n\). In other words, I apply Proposition 20.8 (g) with the observation that \(\Phi\) is the quotient map. So then the \(exp\) map of \(T^n\) is really just quotienting after applying the exponential map

Question 5: Lee 20-6

Part A:Since \(A= e^B\), we can set \(C = e^\frac{B}{2}\). Then \(C^2 = e^B = A\)

Part B: Recall that \(A= \begin{pmatrix} \frac{-1}{2} & 0 \\ 0 & -2 \end{pmatrix}\)

If \(A \in Gl(n,\mathbb{R})\) was in the image, then there would exist a \(C\) such that when squared, is

\(\begin{pmatrix} a & b \\ c & d \end{pmatrix} ^2 = \begin{pmatrix} a^2+bc & b(a+d) \\ c(a+d) & bc+d^2 \end{pmatrix} \)

But this then forces either \(b\) or \(c\) is 0 which is a contradiction, or \(a =-d\) which is a contradiction too.

Question 6: Commuting Exponentials Answer:

Part A: If \([X,Y] =0\), then as vector fields, \(X\) and \(Y\) commute with their flows.

Part B: If \(G\) is Abelian, then the previous proposition immediately implies that \(\frak{g}\) is Abelian too. Since \(exp(tX)exp(sY) \in G\)

If \(\frak{g}\) commutes, then at any point \(p \in G\), if it were to be acted upon by the flow corresponding to a left shift, \(L_q, q\in G\), this corresponds to \(p+q\), but by commutivity of the flows, if we had \(q\) acted upon by the left shift \(L_p\), we get \(q+p\). So then since \([L_q,L_p]=0\) moving by q at p is the same as moving by p at q. Hence, these elements commute.(This is how I am thinking about the problem, not very formal. I'm trying to find the math to put to the words)