Lecture Notes

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===== Tensor Analysis for physics and engineering students =====

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. 

=====  Curvi-linear coordinate =====
** 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 ** 
  - 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  $$ x(r,\theta) = r \cos\theta, \quad y(r,\theta) = r \sin \theta $$
  - 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 $U \subset \R^n$. 



===== Tangent Vector  =====
$$\gdef\b{\mathbf}$$ 
$$\gdef\d{\partial}$$

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

In order to represent both the position and the velocity ((the use of the terminology 'velocity' is not standard in math.)), 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 $\d_\theta$. 


===== Dual Vector Space and Dual basis =====
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 
$$ \langle -, - \rangle: V \otimes V^* \to \R.$$


===== Tensor algebra and Exterior Algebra =====
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. 

==== Volume Forms and Determinant ====
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. 

==== (constant coefficient) Metric Tensor ====
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 
$$ g = \sum_{i,j=1}^n (e_i, e_j) h_i \otimes h_j $$

==== Metric Tensor ====
Lecture Notes

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