1. Let $V = \R^2$, and let $v = (3,2)$ in the Cartesian basis of $\R^2$. Now, we choose another basis as follows $$ e_1 = (2,1), \quad e_2 = (0,1) $$. Expand $v$ in terms of $e_1, e_2$.

2. Let $V = \{a + bt + ct^2 \mid a, b, c \in \R\} $ be the space of polynomials of degree at most 2. Let $f_1(t), f_2(t), f_3(t)$ be three elements in $V$, given as follows $$ f_1(2) = 1, \quad f_1(3) = 0, \quad f_1(5) = 0 $$ $$ f_2(2) = 0, \quad f_2(3) = 1, \quad f_2(5) = 0 $$ $$ f_3(2) = 0, \quad f_3(3) = 0, \quad f_3(5) = 1 $$ Then

• Show that $f_1, f_2, f_3$ forms a basis of $V$.
• Expand $f = 3t^2 + 1$ in terms of this basis.

3. Consider the new coordiante $(u,v)$ on $\R^2$, related to $(x,y)$ by $$ x = \cosh(u) \cos(v), \quad y = \sinh(u) \sin(v) $$

• Find the metric tensor $g$ in terms of $u,v$ (or equivalently, the line element $ds^2$),
• The volume element.
• The expansion of $\d_u$ and $\d_v$ using $\d_x$ and $\d_y$.
• (bonus) If a function on $\R^2$ is given by $f(u,v) = 2u + 3v$, find its gradient expressed using basis vectors $\d_u, \d_v$.
• (bonus) Find the divergence of the vector field $V = \d_u$.

You may use the formula $$ grad(f) = g^{ij}\, \d_i f \,\d_j $$ and $$ div(V^i \d_i) = \frac{1}{\sqrt{g}} \d_i(\sqrt{g} V^i) $$ where we used Einstein summation convention, and $\d_i = \frac{\d}{\d u_i}$, $u_1 = u, u_2 = v$.

1. Draw the picture to see. You should draw a parallogram with $v$ along the diagonal, and two sides parallel to $e_1,e_2$.

2. Since $V$ is 3 dim, suffice to show that $f_i$ are linearly independent. That is, if we have $$ 0 = c_1 f_1(t) + c_2 f_2(t) + c_3 f_3(t) $$ for all $t$, then we need to show $c_i=0$. Plug in $t=2,3,5$ to conclude this is indeed so.

Same thing for the next problem, suppose we try to find the expansion $$ f(t) = c_1 f_1(t) + c_2 f_2(t) + c_3 f_3(t) $$ then we can get $c_i$ by plug-in $t=2,3,5$.

You may worry about: what about $t$ equal to other values? Well, since we have an equation about degree two polynomials, if it holds at 3 points, then it holds for all $t$.

3. This is good place to introduce another trick, the relation about 2-dim orthogonal coordinate and holomorphic functions.

Suppose $z = f(w)$ is a holomorphic function, if we write $z = x+ i y$, $w = u + iv$, we get Cauchy-Riemmann equation $$ \begin{cases} \frac{\d x}{\d u} = \frac{\d y}{\d v} \cr \frac{\d x}{\d v} = - \frac{\d y}{\d u} \end{cases} $$ Hence, we have $$ g(\frac{\d }{\d u} ,\frac{\d }{\d u}) = \left( \frac{\d x}{\d u} \right)^2 + \left( \frac{\d y}{\d u} \right)^2 = \left( \frac{\d y}{\d v} \right)^2 + \left( - \frac{\d x}{\d v} \right)^2= g(\frac{\d }{\d v} ,\frac{\d }{\d v})$$ and $$ g(\frac{\d }{\d u} ,\frac{\d }{\d v}) = \frac{\d x}{\d u} \frac{\d x}{\d v} + \frac{\d y}{\d u} \frac{\d y}{\d v} = - \frac{\d x}{\d u} \frac{\d y}{\d u} + \frac{\d y}{\d u} \frac{\d x}{\d u} = 0 $$ Hence, we have orthogonal coordinate, and furthermore, the length of the two basis vectors are the same $g_{uu} = g_{vv}$.

In this problem, we have $$ x + i y = \cosh(u + iv) $$ so the above method applies. Another examples is $$ y = uv , x = (u^2 - v^2)/2 $$ this is from $$ x + i y = (u + iv)^2/2.$$

Of course, one can do the problem without using the above trick. One then do $$dx = … du + … dv, \quad dy = … du + … dv$$ then plug into $$ ds^2 = dx^2 + dy^2 $$ to express $ds^2$ using $du$ and $dv$. You should get in the end $$ ds^2 = (\cosh^2(u)- \cos^2(v)) (du^2 + dv^2) $$ Or, the factor can be written in different ways, since $$ \cosh^2(u)- \cos^2(v) = \sinh^2(u) + \sin^2(v) .$$

Let $H^2 = \sinh^2(u) + \sin^2(v)$, then we have $$ g = [g_{ij}] = \begin{pmatrix} H^2 & 0 \cr 0 & H^2 \end{pmatrix} $$ So we have $g_{ij} = \delta_{ij} H^2, \quad g^{ij} = H^{-2} \delta_{ij}$.

The volume element is $$ \sqrt{g} du dv = (\sinh^2(u) + \sin^2(v)) du dv$

The expansion is $$ \d_u = \d_u(x) \d_x + \d_u(y) \d_y = … $$

The gradient is $$ grad(f) = H^{-2} \d_u(f) \d_u + H^{-2} \d_v(f) \d_v = H^{-2} (2 \d_u + 3 \d_v) $$

The divergence of $V$ is $$ div(V) = H^{-2} \d_u (H^2) = \frac{2 \cosh(u)\sinh(u)}{ \sinh^2(u) + \sin^2(v) }$$ More details: $V = 1 \d_u + 0 \d_v$ so $V^1 = 1$, $V^2=0$, and $\sqrt{g} = \sqrt{H^4} = H^2$. Plug in the divergence formula will yield the answer.

In Boas' notation, we have $e_1 = \d_u / \| \d_u \| = \d_u / H$, so $$V = \d_u = H e_1$$ and $V^1_{Boas} = H$, also $h_1 = h_2 = H$. Using that Boas formula, we have $$div V = \frac{1}{h_1h_2} \d_u( h_2 V^1_{Boas}) = H^{-2} \d_u (H^2) $$ yield the same answer.