levi-civita symbol
2-d
In two dimensions, the Levi-Civita symbol is defined by: \[ \varepsilon_{i j}=\left\{\begin{aligned} +1 & \text { if }(i, j)=(1,2) \\ -1 & \text { if }(i, j)=(2,1) \\ 0 & \text { if } i=j \end{aligned}\right. \]
3-d
In three dimensions, the Levi-Civita symbol is defined by: \[ \varepsilon_{i j k}=\left\{\begin{aligned} +1 & \text { if }(i, j, k) \text { is }(1,2,3),(2,3,1), \text { or }(3,1,2) \\ -1 & \text { if }(i, j, k) \text { is }(3,2,1),(1,3,2), \text { or }(2,1,3) \\ 0 & \text { if } i=j, \text { or } j=k, \text { or } k=i \end{aligned}\right. \]
n-d
More generally, in \(n\) dimensions, the Levi-Civita symbol is defined by: \[ \varepsilon_{a_1 a_2 a_3 \ldots a_n}=\left\{\begin{aligned} +1 & \text { if }\left(a_1, a_2, a_3, \ldots, a_n\right) \text { is an even permutation of }\\ &(1,2,3, \ldots, n) \\ -1 & \text { if }\left(a_1, a_2, a_3, \ldots, a_n\right) \text { is an odd permutation of }\\&(1,2,3, \ldots, n) \\ 0 & \text { otherwise } \end{aligned}\right. \] some properties \[ \epsilon_{i j k} \epsilon_{i m n}=\delta_{j m} \delta_{k n}-\delta_{j n} \delta_{k m} \]
cross product
In any basis, the cross-product \(a \times b\) is given by the tensorial formula \(E_{i j k} a^i b^j\) where \(E_{i j k}\) is the covariant Levi-Civita tensor (we note the position of the indices). That corresponds to the intrinsic formula given here. Cross product - Wikipedia
The cross product can be defined in terms of the exterior product. It can be generalized to an external product in other than three dimensions. \({ }^{[20]}\) This generalization allows a natural geometric interpretation of the cross product. In exterior algebra the exterior product of two vectors is a bivector. A bivector is an oriented plane element, in much the same way that a vector is an oriented line element. Given two vectors a and \(b\), one can view the bivector \(a \wedge b\) as the oriented parallelogram spanned by \(a\) and \(b\). The cross product is then obtained by taking the Hodge star of the bivector \(a \wedge b\), mapping 2-vectors to vectors: \[ a \times b=*(a \wedge b) \text {. } \]
only valid for three dimensions \[ \star(\mathbf{u} \wedge \mathbf{v})=\mathbf{u} \times \mathbf{v} \quad \star(\mathbf{u} \times \mathbf{v})=\mathbf{u} \wedge \mathbf{v} \]
difference of wedge and cross product
The cross product (blue vector) in relation to the exterior product (light blue parallelogram). The length of the cross product is to the length of the parallel unit vector (red) as the size of the exterior product is to the size of the reference parallelogram (light red).
叉积是纯数学计算，其值是一个向量，楔积是增加维度的计算，经过楔积空间维度变成被计算的两向量维度之和（线性无关），该空间的法向与叉积一致，且两者的数值一致。
I would have left this as a comment, but apparently I can't comment yet.
To answer your final question, why introduce the wedge, the point is that the wedge product (as explained in the Wikipedia article) is a notion that generalizes to \(\mathbb{R}^n\) and indeed any vector space---in general the output is what is called a "bivector". Now, it just so happens that in \(\mathbb{R}^3\) there is a natural identification between bivectors and vectors, and the vector that corresponds to the (bivector) wedge product is indeed the cross product, but this does not work in other vector spaces.
So, to sum up, the notion of cross product is specific to \(\mathbb{R}^3\), whereas wedge product makes sense in any vector space.
Some properties
\[ \begin{aligned} & \vec{u} \times \vec{v}=-\vec{v} \times \vec{u} \\ & \vec{u} \times(\vec{v}+\vec{w})=\vec{u} \times \vec{v}+\vec{u} \times \vec{w} \end{aligned} \]
wedge product and vector
two vector and 2-d
\[ \begin{aligned} \mathbf{v} \wedge \mathbf{w} & =\left(a \mathbf{e}_1+b \mathbf{e}_2\right) \wedge\left(c \mathbf{e}_1+d \mathbf{e}_2\right) \\ & =a c \mathbf{e}_1 \wedge \mathbf{e}_1+a d \mathbf{e}_1 \wedge \mathbf{e}_2+b c \mathbf{e}_2 \wedge \mathbf{e}_1+b d \mathbf{e}_2 \wedge \mathbf{e}_2 \\ & =(a d-b c) \mathbf{e}_1 \wedge \mathbf{e}_2 \end{aligned} \]
three vector and3-d
\[ \small\mathbf{u} \wedge \mathbf{v} \wedge \mathbf{w}=\left(u_1 v_2 w_3+u_2 v_3 w_1+u_3 v_1 w_2-u_1 v_3 w_2-u_2 v_1 w_3-u_3 v_2 w_1\right)\left(\mathbf{e}_1 \wedge \mathbf{e}_2 \wedge \mathbf{e}_3\right) \]
hodge star
Let \(e_i, i=1,2, \ldots, n\), be an orthonormal basis of \(V\) and \(e^i\) the corresponding dual basis of \(V^*\). We define a metric \({g}^*\) on \(V^*\) by declaring this basis to be orthonormal: \[ \stackrel{*}{g}\left(e^i, e^j\right):=\eta_{i j} \] where we have \(\eta_{ij}=e^i\cdot e^j\), then \[ *\left(e^{i_1} \wedge \cdots \wedge e^{i_p}\right):=\varepsilon_{i_1 \cdots i_n} \eta^{i_1 i_1} \cdots \eta^{i_p i_p} e^{i_{p+1}} \wedge \cdots \wedge e^{i_n} \] where \(\eta_{ij}=\frac{1}{\eta^{ij}}\)
:exclamation::exclamation::exclamation: differentiate and analyse
orthonormal basis
e.g. \[ \begin{aligned} & \left\{v_1, v_2, v_3, \ldots, v_n\right\} \\ & \left\|v_i\right\|=1 \end{aligned} \] where \(||a||\) is the modulus of vector
good coordinate system (coordinate basis vectors of a coordinate system) \(\Leftrightarrow\) orthonormal basis
\(d s^2=d \boldsymbol{s} \cdot d \boldsymbol{s}=\left(d x^a \boldsymbol{e}_a\right) \cdot\left(d x^b \boldsymbol{e}_b\right)=\left(\boldsymbol{e}_a \cdot \boldsymbol{e}_b\right) d x^a d x^b .\) and note that here \(g_{a b}=\boldsymbol e_a \cdot \boldsymbol e_b\) and \({\boldsymbol e_a,\boldsymbol e_b}\) aren't coordinate basis vectors (here \(e_a\) is different from the preceding)
\(g_{\mu \mu}\neq 1\)
is metric \(g_{\mu \mu}\) the inner product of orthonormal basis vector?
Not exactly. Here's why:
The metric tensor \(g_{\mu \nu}\) is not the inner product of orthonormal basis vectors. Instead, it is a fundamental object in differential geometry and general relativity that encodes the geometric properties of a space or spacetime.
Metric Tensor
The metric tensor, represented by components \(g_{\mu\nu}\), provides a way to define inner products in a general curved spacetime (or any space with a non-standard notion of distance and angle). It's a generalization of the idea of the dot product in standard Euclidean spaces.
How the Metric Tensor Defines Inner Products
Basis Vectors: In general relativity, you might have coordinate basis vectors (which might not be orthonormal). Let's denote these by \(e_{\mu}\) and \(e_{\nu}\).
Inner Product: The metric tensor defines the inner product between these basis vectors as: \[ g_{\mu\nu} = e_{\mu}\cdot e_{\nu} \]
Key Differences from Orthonormal Bases
- Not Unit Vectors: The basis vectors \(e_{\mu}\) don't have to have unit magnitude. The metric tensor components will adjust to account for their lengths.
- Not Necessarily Orthogonal: The basis vectors don't have to be perpendicular, but if in orthogonal coordinate they are perpendicular The metric tensor components will account for any angles between them as well.
Important Point: While the diagonal elements of the metric tensor (\(g_{\mu\mu}\)) do relate to the squared magnitudes of the corresponding basis vectors, the key idea is that the metric tensor defines the inner product between any two basis vectors, not just along diagonal elements.
In spherical coordinate systems, the diagonal elements of the metric tensor are not equal to 1 because the metric tensor encodes the geometry of the space, and the geometry of a sphere is different from that of a flat space where the diagonal elements would be 1 .
The line element in spherical coordinates \((r, \theta, \phi)\) is given by: \[ d s^2=d r^2+r^2 d \theta^2+r^2 \sin ^2 \theta d \phi^2 \]
Here, \(d s^2\) represents the square of the proper distance (or spacetime interval) between two infinitesimally close events. The metric tensor components can be derived from this line element and are given by: \[ \begin{aligned} g_{r r} & =1 \\ g_{\theta \theta} & =r^2 \\ g_{\phi \phi} & =r^2 \sin ^2 \theta \end{aligned} \]
As we can see, the diagonal elements \(g_{r r}, g_{\theta \theta}\), and \(g_{\phi \phi}\) are not all equal to 1 . This is because the geometry of the spherical coordinate system introduces curvature, especially in the \(\theta\) and \(\phi\) directions.
The \(g_{r r}\) component is 1 , which corresponds to the radial coordinate, and it represents the flat geometry along the straight lines in the radial direction. However, the angular components \(g_{\theta \theta}\) and \(g_{\phi \phi}\) are multiplied by \(r^2\) and \(r^2 \sin ^2 \theta\), respectively, to account for the curvature of the sphere.
The factor of \(r^2\) in \(g_{\theta \theta}\) and \(g_{\phi \phi}\) reflects the fact that the circumference of a circle on the surface of a sphere decreases as one moves from the equator towards the poles. This is a direct consequence of the spherical geometry, and it is why the diagonal elements of the metric tensor in spherical coordinates are not all equal to 1 .
In summary, the non-unity values of the diagonal elements of the metric tensor in spherical coordinates are a reflection of the curved geometry of the sphere, which differs from the flat geometry that would be described by a metric tensor with all diagonal elements equal to 1 . In Cartesian coordinates, moving a unit distance in any coordinate direction always corresponds to the same spatial displacement, which is not the case in spherical coordinates due to the curvature of the coordinate lines. The metric tensor thus encodes how the geometry of space itself changes with position in this coordinate system。
Orthonormal basis vectors at a point
At any given point \(P\) in a manifold, it is often useful to define a set of orthonormal basis vectors \(\hat{\boldsymbol{e}}_a\) in \(T_P\), which are chosen to be of unit length and orthogonal to one another. This is expressed mathematically by the requirement that at \(P\) \[ \hat{\boldsymbol{e}}_a \cdot \hat{\boldsymbol{e}}_b=\eta_{a b} \] where \(\left[\eta_{a b}\right]=\operatorname{diag}( \pm 1, \pm 1, \ldots, \pm 1)\) is the Cartesian line element of the tangent space \(T_P\) and depends on the signature of the (in general) pseudo-Riemannian manifold (see Section 2.13). These orthonormal basis vectors need not be related to any particular coordinate system that we are using to label the manifold, although they can always be defined by, for example, giving their components in a coordinate basis. Moreover, it is clear from (3.7) that the orthonormal basis vectors \(\hat{\boldsymbol{e}}_a\) at \(P\) are in fact the coordinate basis vectors of a coordinate system for which \(g_{a b}(P)=\eta_{a b}\) (or \(g_{a b}(P)=\delta_{a b}\) for a strictly Riemannian manifold).