Homework 2 Problem 3


Consider $M = \mathbb{T}^2$ with local coordinates $(\theta,\phi)$ and $N = \mathbb{S}^3$ with local coordinates $(a,b,c)$.


\begin{align} \left\langle \negmedspace \left\langle u, w \right\rangle \negmedspace \right\rangle= \langle \negmedspace \langle u_\theta \frac{\partial}{\partial \theta} + u_\phi \frac{\partial}{\partial \phi}, w_\theta \frac{\partial}{\partial \theta} + w_\phi \frac{\partial}{\partial \phi} \rangle \negmedspace \rangle = 5 u_\theta w_\theta + u_\theta w_\phi + u_\phi w_\theta + 2 u_\phi w_\phi \end{align}

for all $u, w \in T_q M$.
Compute $v^\flat$ with respect to this inner product.


We have the relationships,

\begin{align} \langle \negmedspace \langle \alpha^\# , w \rangle \negmedspace \rangle = & ~\langle \alpha , w \rangle \end{align}
\begin{align} (v^\flat)^\# = &~ v \end{align}

$\forall v,w \in T_qM \text{ and } \alpha \in T_q^*M$. Let

\begin{align} v &= v_\theta\frac{\partial}{\partial \theta} + v_\phi \frac{\partial}{\partial \phi}&\text{, and}\\ v^\flat &= v^\theta d\theta + v^\phi d\phi \end{align}

We want to compute $v^\theta$ and $v^\phi$. Using Equation (3) we get

\begin{align} (v^\flat)^\# &= (v^\theta d\theta + v^\phi d\phi)^\# = v_\theta\frac{\partial}{\partial \theta} + v_\phi \frac{\partial}{\partial \phi} = v \end{align}

Let $\alpha = v^\flat$. Then $\alpha^\# = v$. Rewriting Equation (2),

\begin{align} \langle \negmedspace \langle \alpha^\# , w \rangle \negmedspace \rangle =& \langle \alpha,w\rangle \end{align}

Computing both sides of above equation we get,

\begin{align} \langle \alpha,w\rangle &= \langle v^\theta d\theta + v^\phi d\phi , w_\theta \frac{\partial}{\partial \theta} + w_\phi \frac{\partial}{\partial \phi} \rangle \\&= v^\theta w_\theta + v^\phi w_\phi \end{align}
\begin{align} \langle \negmedspace \langle \alpha^\# , w \rangle \negmedspace \rangle &= \langle \negmedspace \langle v , w \rangle \negmedspace \rangle = \langle \negmedspace\langle v_\theta \frac{\partial}{\partial \theta} + v_\phi \frac{\partial}{\partial \phi},w_\theta \frac{\partial}{\partial \theta} + w_\phi \frac{\partial}{\partial \phi} \rangle \negmedspace\rangle \\ &= 5 u_\theta w_\theta + u_\theta w_\phi + u_\phi w_\theta + 2 u_\phi w_\phi \\ &= (5v_\theta + v_\phi)w_\theta + (v_\theta + 2v_\phi)w_\phi \end{align}

Comparing coefficients of $w_\theta$ and $w_\phi$ in Equation (7) and Equation (8), we have

\begin{align} v^\theta =& 5v_\theta + v_\phi \\ v^\phi =& v_\theta + 2v_\phi \end{align}


\begin{align} v^\flat = v^\theta d\theta + v^\phi d\phi = (5v_\theta + v_\phi )d\theta + (v_\theta + 2v_\phi) d\phi \end{align}


Tightly-spaced double angle brackets can be created by inserting negative space between single angle brackets, as in the problem description above.

[kcrane] I think there's a much simpler explanation for this problem. All you want to do is find a 1-form that acts like it's taking an inner product with the vector field $v$ whenever it eats a vector $w$. You're given that an inner product with $v$ looks like this:

\begin{align} 5 v_{\theta} w_{\theta} + v_{\theta}w_{\phi} + v_{\phi}w_{\theta} + 2 v_{\phi}w_{\phi} \end{align}

and you know that a 1-form $\alpha$ looks like this whenever it's eating a vector $w$:

\begin{align} \alpha = \alpha_{\theta} w_{\theta} + \alpha_{\phi} w_{\phi}. \end{align}

To get these two things to be equal, all you have to do is look at the four terms in the inner product and decide that

\begin{align} \alpha_{\theta} = 5v_{\theta} + v_{\phi} \end{align}


\begin{align} \alpha_{\phi} = v_{\theta} + 2v_{\phi}. \end{align}

(since those are the only values for the 1-form that make it look like the inner product!).
So then $v^{\flat} = \alpha = \left(5v_{\theta} + v_{\phi}\right) \mathrm{d}_{\theta} + (v_{\theta} + 2v_{\phi}) \mathrm{d}_{\phi}$, just like you had in the previous solution.

Scott agrees!

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