This is for HW 8, my main regeneration

Question 1: Round Metric

This was shown in Lee's Riemannian Manifolds: An Introduction to Curvature. Lemma 3.4

Question 2: Euclidean geodesics

We compute \(\frac{d}{ds} L_{\bar{g}}(\gamma_s) = \frac{d}{ds} (\int_{0}^{1} |\gamma_{s}'(t)|_{\bar{g}} dt)\)

Expanding this yields:

\(\begin{align} \frac{d}{ds}(\int_{0}^{1}|\gamma'(t) +s\tilde{\gamma}'(t) -s\gamma'(t)|_{\bar{g}}) &= \frac{d}{ds}(\int_{0}^{1}|(1-s)(y-x) + s\tilde{\gamma}'(t)|_{\bar{g}} dt) \\ &= \int_{0} ^{1} |(x-y) + \tilde{\gamma}'(t)|_{\bar{g}} dt \\ &= \int_{0}^{1} \sqrt{((y-x)+\tilde{\gamma}'(t))^2} dt \\ &= t(y-x) + \tilde{\gamma}(t) |_{0}^{1} \end{align} \)

The third inequality comes from differentiating under the integral since they are different variables.

What the last result shows us is that the only way this new path would be any improvement to the usual Euclidean distance is if \(\tilde{\gamma}\) were 0. Thus, we see that Euclidean geodesics are exactly straight lines.

Question 3: Complete metrics on \(\mathbb{R}\)

Assume that the integrals converge (say to some point \(p\in \mathbb{R}\)). We show the following for the positive case, the negative integral case is very similar. If this space is complete, then every cauchy sequence converges. Then, consider the sequence \(0 \rightarrow t_1 \rightarrow t_2 \rightarrow \ldots \rightarrow \infty\) so that then \(\int_{0} ^{\infty} \sqrt{f(t)}dt = \sum_{i=0} ^{\infty}\int_{t_i}^{t_{i+1}}\)

We then consider a cauchy subsequence, \(\{k_{j}\}\) that approaches point \(p\) Then we see that we can approximate the \(k_j\) by a sequence of sums of integrals with bounds \(\{t_{n_i}, t_{n_i +1} \}\). Thus we have a Cauchy sequence of sums of integrals that approach the bound.

I didn't really understand how our conversation led to showing that this sequence contradicted the Cauchy property of the sequence. I guess the point is that then if you take some open set around \(p\) and then compute the distance between \(p\) and a point to the right of it. The triangle inequality just blows up in your face since then \(d_g(0, p) \geq d_g(0, \infty) + d_g(p,q)\)

I attached some of my scratchwork on paper and maybe you can point out in the math where I am not thinking about things (often haha)

Question 4: Symmetric connections

Part A: We want to show that \(\tau (fX_1 +gX_2, Y) = f \tau(X_1,Y) + g\tau(X_2,Y)\) where \(f,g \in C^{\infty}(M)\)

Observe that the left hand side is:

\(\nabla_{fX_1 +gX_2} Y - \nabla_{Y}(fX_1 +gX_2) - [fX_1 +gX_2, Y] \)

\(= f \nabla_{X_1}Y +g \nabla_{X_2}Y - (f \nabla_{Y}X_1 + g\nabla_{Y}X_2 + (Yf)X_1 + (Yg)X_2 - ( (fX_1 +gX_2)Y^j -Y(fX_1 +gX_2)^j) \frac{\partial}{\partial x^j} \)

\(= (f \nabla_{X_1}Y - f\nabla_{Y}X_1 -(Yf)X - (fX_1 Y^j - Y(fX_1)^j)\frac{\partial}{\partial x^j} + (g \nabla_{X_2}Y - g\nabla_{Y}X_2 -(Yg)X_2 - (gX_2 Y^j - Y(gX_2)^j)\frac{\partial}{\partial x^j} \)

\(= f \tau(X_1,Y) +g\tau(X_2,Y)\)

Part B: Here we chase down some definitions. Let \(X = X^i \partial_i\), and \(Y = Y^j \partial^j\). So then we see that \(\nabla_X Y - \nabla_Y X - [X,Y] = X^i(Y^j \Gamma_{ij}^k \partial_k + Y_i ^j + \partial_j)- Y^j(X^i \Gamma_{ji}^k \partial_k + X^{i}_{j} \partial_i) - (X^i Y^j_i -Y^i X^j_i)\partial_J\)

Question 5: Space connections

We want to show that the difference tensor field \(A: \frak{X} (M) \times \frak{X}(M) \rightarrow \frak{X}(M)\) defined by the map

\(\begin{equation} A(X,Y) = \nabla_X ^{1}Y - \nabla_X ^{0} Y \end{equation}\)

has all the nice properties of a connection. So we'll show that now.

Linear Product Rule: We want to show that \(A(fX_1+gX_2,Y) = fA(X_1,Y)+gA(X_2,Y)\) Observe that the left hand side is exactly \(\begin{split} \nabla_{fX_1+gX_2} ^{1} Y - \nabla_{fX_1+gX_2}^0 Y &= (f \nabla_{X_1}^1 Y + g\nabla_{X_2^{1}}Y) - (f \nabla_{X_1}^{0}Y + g\nabla_{X_2}^{0}Y) \\ &= f(\nabla_{X_1}^{1} Y - \nabla_{X_1}^{0}Y ) + g(\nabla_{X_2}^{0} Y - \nabla_{X_2}^{0}Y) \\ &= f(A(X_1,Y)) +g(A(X_2,Y)) \end{split}\)

Next, if \(a,b \in \mathbb{R}\), we wish to that \(A(X, aY_1 +bY_2) = a(A(X,Y_1)) + b(A(X,Y_2))\)

Manipulating the left hand side, we see that:

\(\begin{align} \nabla_{X}^{1}(aY_1+bY_2) - \nabla_{X}^{0}(aY_1 +bY_2) &= a \nabla_{X}^{1}Y_{1} + b \nabla_{X}^{1}Y_{2} - a\nabla_{X}^{0}Y_{1} - b\nabla_{X}^{0}Y_{2} \\ &= a(\nabla_{X}^{1}Y_{1} - \nabla_{X}^{0}Y_{1}) + b(\nabla_{X}^{1}Y_{2} - \nabla_{X}^{0}Y_{2}) \\ &= a(A(X,Y_1)) + b(A(X,Y_2)) \end{align}\)

Finally we show that if \(f \in C^{\infty}(M)\) then \(A(X,fY) = f(A(X,Y))\)

The left hand side is equal to:

\(\begin{align} \nabla_{X}^{1}(fY) - \nabla_{X}^{0}(fY) &= f \nabla_{X}^{1}(Y) + (Xf)Y - f \nabla_{X}^{0}Y - (Xf)Y \\ &= f A(X,Y) \end{align}\)