math121b:02-07
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math121b:02-07 [2020/02/07 00:04] pzhou |
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| $$g_{ij} = g (\frac{\d}{\d u_i}, \frac{\d}{\d u_j}) = \sum_k \frac{\d x_k}{\d u_i} \frac{\d x_k}{\d u_j} $$ | $$g_{ij} = g (\frac{\d}{\d u_i}, \frac{\d}{\d u_j}) = \sum_k \frac{\d x_k}{\d u_i} \frac{\d x_k}{\d u_j} $$ | ||
| - | If we view $g_{ij}$ as entries of a matrix $[g_{ij}]$, then we denote $g^{ij}$ the entries of the inverse matrix. Let $|g| = \det( [g_{ij}] )$ be the absolute value of the determinant of the matrix. Then, our formula for the Laplacian operator is (where $\d_i = \d / \d u_i$) | + | If we view $g_{ij}$ as entries of a matrix $[g_{ij}]$, then we denote $g^{ij}$ the entries of the inverse matrix. Let $|g| = \det( [g_{ij}] )$ be the determinant of the matrix. Then, our formula for the Laplacian operator is (where $\d_i = \d / \d u_i$) |
| $$ \Delta f = \sum_{i,j} \frac{1}{\sqrt{|g|}} \d_i (g^{ij} \sqrt{|g|} \d_j(f)) $$ | $$ \Delta f = \sum_{i,j} \frac{1}{\sqrt{|g|}} \d_i (g^{ij} \sqrt{|g|} \d_j(f)) $$ | ||
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| $$ \Delta f = \frac{1}{\sqrt{|g|}}\sum_{i, | $$ \Delta f = \frac{1}{\sqrt{|g|}}\sum_{i, | ||
| + | ===== Tying up the loose ends ===== | ||
| + | We will prove the two claims above. The first is about volume element. | ||
| + | ==== Volume Element ==== | ||
| + | I will motivate this using vector space. Suppose we are in $\R^n$, and we have a new basis $\gdef\t\tilde \t e_1, \cdots, \t e_n$. We are interested in getting the volume spanned by the skewed cube with sides given by these basis vectors. | ||
| + | |||
| + | You say, this is easy! Suppose we know the components of $\t e_i$ in the standard basis, then we can write | ||
| + | $$ \t e_i = \sum_j a_{ij} e_j $$ | ||
| + | and $| \det a_{ij} |$ is our desired volume. | ||
| + | |||
| + | The fancy way of saying this is that | ||
| + | $$ \det a_{ij} = \frac{\t e_1 \wedge \cdots \wedge \t e_n }{e_1 \wedge \cdots \wedge e_n} $$ | ||
| + | We take the volume of $e_1 \wedge \cdots \wedge e_n$ to be one. Then take the absolute number of $\det a_{ij}$, just in case it is a negative number. | ||
| + | |||
| + | However, we are not given $a_{ij}$! We are only given the $g_{ij}$, which are | ||
| + | $$ g_{ij} = g(\t e_i, \t e_j) $$ | ||
| + | Fortunately, | ||
| + | Concretely, we have | ||
| + | $$ g_{ij} = \sum_k a_{ik} a_{jk} $$ | ||
| + | Hence, if you view $G$ as the matrix with entry $g_{ij}$ and $A$ as a matrix with entries $a_{ij}$, we get | ||
| + | $$G = A A^T \Rightarrow \det G = \det (A A^T) = \det A \cdot \det A^T = (\det A)^2$$ | ||
| + | Thus, to get $|\det A|$, we can just do | ||
| + | $$ |\det A| = \sqrt{\det G} $$ | ||
| + | and don't worry, $G$ is positive definite, hence $\det G > 0$, so the square-root make sense. | ||
| + | |||
| + | ==== Cotangent Vector ==== | ||
| + | For $p \in \R^n$, we define $T_p \R^n$ to be the dual vector space of $T_p \R^n$, and an element of it is called a // | ||
| + | |||
| + | If $f$ is a function, then $df(p)$ is an element of $T^*_p \R^n$. In Cartesian coordiante, we can write | ||
| + | $$ df = \sum_i | ||
| + | |||
| + | Actually, if we are given any coordinate system $(u_1, \cdots, u_n)$, then we have a basis of $T^*_p \R^n$, given by $du_1, \cdots, d u_n$. And any $df$ can be written as | ||
| + | $$ df = \sum_i \frac{\d f}{\d u_i} d u_i. $$ | ||
| + | |||
| + | ==== Inner product of Covector ==== | ||
| + | But, you wrote $ (df, d\varphi)_g$, | ||
| + | |||
| + | Wait, $g$ is only inner product on $T_p \R^n$, how do you get inner product on $T^*_p \R^n$. | ||
| + | |||
| + | Well, in general, if $V$ has an inner product, we can equip $V^*$ with an inner product in the following way. Take $e_i$ an orthonormal basis of $V$, and $h^i$ the dual basis of $V^*$. We declare $h^i$ to be an orthonormal basis of $V^*$. This declaration defines an inner product on $V^*$, sometimes denoted as $g^*$. | ||
| + | |||
| + | That is still kind of abstract, let us use basis. Say, you pick a basis $e_i$ of $V$, and you know that | ||
| + | $g_{ij} = g(e_i, e_j)$. You want to find out $g^*$ on $V^*$ using the dual basis $h^i$. We then know that | ||
| + | $g^{ij} : = g^*(h^i, h^j)$ forms a matrix, and it is the inverse matrix for the one formed by $g_{ij}$. | ||
| + | (Warning, we do not have $g^{ij} = \frac{1}{g_{ij}$, | ||
| + | |||
| + | That's it. Hence, we write | ||
| + | $$ (df, d\varphi)_g = g^*( \sum_i \frac{\d f}{d u_i} d u_i, \sum_j \frac{\d \varphi}{d u_j} d u_j) = \sum_{i,j } \frac{\d f}{d u_i}\frac{\d \varphi}{d u_j} g^* (d u_i, d u_j) =\sum_{i,j } \frac{\d f}{d u_i}\frac{\d \varphi}{d u_j} g^{ij} | ||
math121b/02-07.1581062674.txt.gz · Last modified: 2020/02/07 00:04 by pzhou
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