Lecture Notes

There are two ways to introduce Laplacian in curvilinear coordinate, one is following Boas, one is following the standard math language. I will present both approaches.

Definition(Smooth functions)

• Let $U \subset \R^n$ be an open set, and $f: U \to \R$ be a function. We say $f$ is a smooth function, if all the partial derivatives of $f$ exists and is smooth, that is, for all $a_1, \cdots, a_n \geq 0$, we have $$ \gdef\d{\partial} \frac{\d^{a_1}}{\d x_1^{a_1}} \cdots \frac{\d^{a_n}}{\d x_n^{a_n}} f \text{ exists and is smooth}$$.
• Let $U \subset \R^n$ be an open set, and $f: U \to \R^k$ be a function, i.e., $$f(x_1,\cdots, x_n) = \left( f_1(x_1, \cdots, x_n),\cdots, f_k(x_1, \cdots, x_n) \right), \quad \text{ for } (x_1, \cdots, x_n) \in U.$$ We say $f$ is a smooth function, if each $f_i$ is a smooth function.

Definition(Curviliear coordinates) Let $U$ be an open set in $\R^n$. A curvilinear coordinate on $U$ is a smooth function $f=(f_1,\cdots,f_n): U \to \R^n$, such that

• $f$ is a bijection between $U$ and its image $f(U)$
• and the inverse function $f^{-1}: f(U) \to U$ is also smooth.

Example

1. Polar coordinate on $\R^2$. Let $U = \R^2 - \{(x,y) | x \leq 0, y = 0\}$ be an open subset in $\R^2$. Then $$ f=(r, \theta) : U \to (0, \infty) \times (-\pi, \pi), \quad (x,y) \mapsto (\sqrt{x^2+y^2}, \theta(x,y)) $$ is an example of curvilinear coordinates, where $\theta(x,y)$ is the rotation angle from the direction of the positive $x$ axis to the ray passing through $(x,y)$. The inverse is given by $$ x(r,\theta) = r \cos\theta, \quad y(r,\theta) = r \sin \theta $$
2. Polar coordinate on $\R^3$. Let $U = \R^3 - \{(x,y,z)| x \leq 0, y=0\}$. We have $$ f: U \to (0, \infty) \times (-\pi, \pi) \times (0, \pi), \quad (x,y,z)\mapsto (r, \theta, \phi) $$ whose inverse is given by $$ f^{-1}: (r, \theta, \phi) \mapsto (x,y,z), \quad (x = r \sin(\phi) \cos\theta, \quad y = r \sin(\phi) \sin(\theta), \quad z = r \cos(\phi). $$

Remark The reason that we cannot take $U$ to be the entire $\R^2$ or $\R^3$ is because we want to have our coordinate map to be a bijection.

Notation We reserve the notation $(x_1, \cdots, x_n)$ to be the standard Cartesian coordinate on $\R^n$. We use notation of a pair $(U, (u_1, \cdots, u_n))$, or $(U, (f_1, \cdots, f_n))$ for a coordinate on $U \subset \R^n$.

$$\gdef\b{\mathbf}$$ $$\gdef\d{\partial}$$

Consider the n-dimensional Euclidean space $\R^n$ with basis vectors $\b e_1, \cdots, \b e_n$. A vector $\b v = v^1 \b e_1 + \cdots + v^n \b e_n$ has two possible meansings

1. it can represent a location in the space $\R^n$. You cannot add two locations (can you add New York to San Francisco?)
2. it can represent a velocity vector (an arrow with direction and length).

In order to represent both the position and the velocity ¹⁾, we need consider the notion of a tangent vector on $\R^n$.

Definition (Tangent vector) A tangent vector on $\R^n$ is a pair $(\b a, \b v)$ representing the location and velocity of a particle, where $\b a \in \R^n$ represent the position, and $\b v \in \R^n$ represent the velocity. The set of tangent vectors with the same position, say equal to $\b a_0$, forms the tangent space at $\b a_0$, $$\R^n|_{\b a_0} = \{ (\b a, \b v) \mid \b a = \b a_0, \b v \in \R^n \}.$$

Warning: Only tangent vectors standing over the same position can be added or subtracted.

Definition (Vector field) A vector field on $U \subset \R^n$ is an assignment of tangent vectors $\b v$ to each point $\b a \in U$, such that $\b v$ varies smoothly with respect to $\b a$.

Example(the vector field $\d / \d u_i$) Let $(U, (u_1, \cdots, u_n))$ be a coordinate system on $U$. Let $p \in U$ be a point, and choose an $i \in \{1, \cdots, n\}$. We will define a tangent vector $\frac{\d}{\d u_i} \vert_p$ at $p$ “physically” as follows. Consider the motion of a particle on $U$, describe by the following curve $\gamma: (-\epsilon, +\epsilon) \to U$, such that $\gamma(0) = p$, and for $t \in (-\epsilon, +\epsilon)$ $$ u_j(\gamma(t)) = \begin{cases} u_j(\gamma(0)) & j \neq i \cr u_j(\gamma(0))+t & j = i \end{cases} $$. Then, we define $\frac{\d}{\d u_i} \vert_p$ to be the velocity of the particle at the moment $t=0$. As we vary $p \in U$, the tangent vectors $\frac{\d}{\d u_i} \vert_p$ forms a vector field, denoted as $\frac{\d}{\d u_i}$.

We sometimes denote $\d / \d u_i$ by $\d_{u_i}$.

Example(Polar coordinate $(r,\theta)$) Draw the picture for $\d_r$ and $\d_\theta$.

Let $V$ be a finite dimensional vector space over $\R$. We let $V^*$ denote the set of linear functions on $V$. One can verify that $V^*$ is also a vector space over $\R$. If $\dim V=n$, then $\dim V^*=n$ as well.

Let $e_1, \cdots, e_n$ be a basis of $V$, to specify an element in $V^*$, we just need to specify its value on the basis elements. We define the following elements $h_1, \cdots, h_n$ in $V^*$: $$ h_i (e_j) = \delta_{ij} $$ One can show that $h_i$ forms a basis of $V^*$. $\{h_i\}$ is said to be the dual basis of $\{e_i\}$.

There is a canonical pairing between $V$ and $V^*$, denoted as $$ \langle -, -\rangle: V \times V^* \to \R, \quad (v, h) \mapsto h(v) $$. It is linear in both $V$ and $V^*$, hence we can extend it to a map of $$ \langle -, - \rangle: V \otimes V^* \to \R.$$

Let $V$ be a finite dimensional vector space over $\R$. We denote the $k$ copies tensor product $V \otimes \cdots \otimes V$ as $V^{\otimes k}$, its elements are linear combinations of terms like $v_1 \otimes \cdots v_k$.

$$\gdef\ot\otimes$$

Definition (Tensor Algebra $T(V)$ ) $$T(V) = \R \oplus V \oplus V^{2} \oplus \oplus V^{3} \oplus \cdots $$ Given two elements $T = w_1 \ot \cdots \ot w_k$ and $T' = v_1 \ot \cdots \ot v_l$, their products is defined by juxtapostion. $$ T \ot T' = w_1 \ot \cdots \ot w_k \ot v_1 \ot \cdots \ot v_l $$

Definition (Exterior product $\wedge^k(V)$) The $k$-th exterior product $\wedge^k(V)$ is the vector space consisting of linear combinations of the following terms $v_1 \wedge \cdots \wedge v_k$, where the expression is linear in each slot, $$ c \cdot (v_1 \wedge \cdots \wedge v_k) = (c v_1) \wedge v_2 \wedge \cdots \wedge v_k $$ $$ (v_1+v_1') \wedge \cdots \wedge v_k = v_1 \wedge \cdots \wedge v_k + v_1'\wedge \cdots \wedge v_k$$ and the expression changes signs if we swap any two slots $$ v_1 \wedge \cdots \wedge v_i \wedge \cdots \wedge v_j\wedge \cdots \wedge v_k = - v_1 \wedge \cdots \wedge v_j \wedge \cdots \wedge v_i\wedge \cdots \wedge v_k, \forall 1 \leq i < j \leq k. $$

If we choose a basis $e_1, \cdots, e_n$ of $V$, then for $0 \leq k \leq n$, the space $\wedge^k(V)$ has a basis consisting of the following vectors $$ e_{i_1} \wedge \cdots \wedge e_{i_k}, \quad 1 \leq i_1 < i_2 < \cdots < i_k \leq n. $$ The basis vectors are labelled by size $k$ subset $I$ of $\{1, \cdots, n\}$, hence we also denote the above basis vector by $e_I$.

We may consider all $\wedge^k V$ together, as $$ \wedge V = \bigoplus_{k=0}^{\dim V} \wedge^k V, \quad \text{ where } \wedge^0 V:= \R.$$ Then, we can define wedge products on $\wedge V$. If $A = v_1 \wedge \cdots \wedge v_k \in \wedge^k V$, $B = w_1 \wedge \cdots \wedge w_l \in \wedge^l V$, then we define the product by juxtaposition $$ A \wedge B := v_1 \wedge \cdots \wedge v_k \wedge w_1 \wedge \cdots \wedge w_l \in \wedge^{k+1} V.$$

Example on $V=\R^3$ Consider the $\wedge^2 V$, its dimension is ${3 \choose 2} = 3$. If we use the standard basis $e_1, e_2, e_3$ on $V$, then we have the following basis for $\wedge^2 V$: $$ e_1 \wedge e_2, \quad e_1 \wedge e_3, \quad e_2 \wedge e_3. $$

For $\wedge^3 V$, it is one-dimensional, with $e_1 \wedge e_2 \wedge e_3$ as a basis.

There is a bijection from $\wedge^2 V \to V$, called “Hodge star” $\star$, which goes as follows: $$ \star: e_1 \wedge e_2 \mapsto e_3, \quad e_2 \wedge e_3 \mapsto e_1, \quad e_3 \wedge e_1 \mapsto e_2. $$

Thus, we may recover our familiar cross-product $\b v \times \b w$ formula as following $$ V \times V \xrightarrow{\wedge} \wedge^2 V \xrightarrow{\star} V. $$

Exercise: convince yourself that $\b v \wedge \b w = \star(\b v \wedge \b w)$.

Remark : If $V=\R^3$, then elements of $\wedge^2 V$ are called pseudo-vectors, and elements of $\wedge^3 V$ are called pseudo-scalar.

Still, in the example of $V=\R^3$. Given three vectors $v_1, v_2, v_3$, how to compute the signed volume formed by the parallelogram $P(v_1, v_2, v_3)$ (skewed boxes) with sides $v_1, v_2, v_3$?

From vector calculus, we know the answer is the determinant of the $3$ by $3$ matrix, whose column-vectors are $v_1, v_2, v_3$. $$ \text{ Volume of } P(v_1, v_2, v_3) = \det \begin{pmatrix} v_{11} & v_{21} & v_{31} \cr v_{12} & v_{22} & v_{32} \cr v_{13} & v_{23} & v_{33} \end{pmatrix} $$

Now, we have another way to express it. $$ \text{ Volume of } P(v_1, v_2, v_3) = \frac{ v_1 \wedge v_2 \wedge v_3}{e_1 \wedge e_2 \wedge e_3} $$ Indeed, since both the numerator and denominators are elements of the one-dim vector space $\wedge^3 V$, the raio makes sense.

Let $V$ be an $n$-dimensional vector space, assume that we have an inner product, i.e., a symmetric bilinear form $$ (-, -): V \times V \to \R $$ such that for any $v \in V \neq 0$, $(v,v) > 0$. Such a space is called an $n$-dimensional Euclidean vector space .

The metric tensor $g$ for $V$ is a rank-2 symmetric tensor $g \in V^* \otimes V^*$. Just as $V^*$ is defined as linear function from $V \to \R$, we may interpret $V^* \otimes V^*$ as bilinear functions $V \times V \to \R$. Under this interpretation, the metric tensor $g$ is the inner product $(-,-)$.

If $e_1,\cdots, \e_n$ are a ortho-normal basis of $V$, and $h_1, \cdots, h_n$ are the dual basis. Then we may write $g$ as $$ g = \sum_{i=1}^n h_i \otimes h_i. $$

In general, for any basis $e_1, \cdots, e_n$ and corresponding dual basis $h_1, \cdots, h_n$, we have $$ g = \sum_{i,j=1}^n (e_i, e_j) h_i \otimes h_j $$