Unit 3: Metric Tensors, Riemannian Spaces & Index Manipulation
Foundations of Riemannian geometry and metric tensor calculus: the fundamental quadratic differential form ds^2 = g_ij dx^i dx^j, Riemannian vs Pseudo-Riemannian manifolds, the covariant metric tensor g_ij and contravariant reciprocal metric tensor g^ij, metric determinant g and its derivative identities, natural basis vectors e_i and dual covectors e^i, the musical isomorphisms for raising and lowering indices (associated tensors), vector magnitudes, Riemannian angles, and orthogonal coordinate systems.
§3.1 The Fundamental Metric Form $ds^2 = g_{ij} dx^i dx^j$ & Riemannian Manifolds
1. The Riemannian Metric Axioms
In 1854, Bernhard Riemann generalized Euclidean geometry to arbitrary $n$-dimensional curved manifolds by postulating that the infinitesimal distance $ds$ between two neighboring points $P(x^i)$ and $Q(x^i + dx^i)$ is given by the square root of a quadratic differential form.
Definition 3.1 (Fundamental Metric Form): A differentiable manifold $V_n$ is a Riemannian space $R_n$ if it is equipped with a symmetric, second-rank covariant tensor field $g_{ij}(x) = g_{ji}(x)$ called the metric tensor, such that the arc length element $ds$ satisfies:
In a proper (positive-definite) Riemannian space, $ds^2 > 0$ for any non-zero displacement $dx^i \ne 0$.
Signature Classifications:
1. Positive-Definite Riemannian Space: All eigenvalues of the matrix $(g_{ij})$ are strictly positive ($++++\dots$). Distance $ds$ is always real and positive.
2. Pseudo-Riemannian (Lorentzian) Space: The metric is non-degenerate ($\det(g) \ne 0$) but indefinite. In 4D relativistic spacetime, the metric signature is $(-, +, +, +)$ (or $(+, -, -, -)$):
where intervals are classified into timelike ($ds^2 < 0$), spacelike ($ds^2 > 0$), and lightlike / null ($ds^2 = 0$).
2. Transformation Law of the Metric Tensor
Let $x^i \to \bar{x}^k$ be a regular coordinate transformation. Since the arc length $ds$ is an intrinsic physical distance, the quadratic form is an absolute scalar invariant:
Substituting the differential transformation $dx^i = \frac{\partial x^i}{\partial \bar{x}^k} d\bar{x}^k$ and $dx^j = \frac{\partial x^j}{\partial \bar{x}^l} d\bar{x}^l$:
Because the coordinate differentials $d\bar{x}^k$ are arbitrary, we deduce:
This rigorously confirms that $g_{ij}$ transforms as a symmetric covariant tensor of rank 2.
§3.2 The Metric Tensor $g_{ij}$, Conjugate Metric $g^{ij}$ & Determinant $g$
1. The Conjugate (Reciprocal) Metric Tensor $g^{ij}$
Let $g = \det(g_{ij})$ denote the determinant of the $n \times n$ matrix of metric components. For a non-degenerate Riemannian space, $g \ne 0$.
Definition 3.2 (Conjugate Metric Tensor): The conjugate (reciprocal) metric tensor $g^{ij}$ is defined as the matrix inverse of $g_{ij}$:
In terms of the matrix cofactors of $g_{ij}$:
where $G^{ij}$ is the cofactor of the entry $g_{ij}$ in the determinant $g$.
By symmetry of $g_{ij}$, the conjugate metric tensor is also symmetric: $g^{ij} = g^{ji}$.
2. Derivative Identities of the Metric Determinant
The derivative of the metric determinant with respect to coordinate $x^k$ plays a central role in constructing Christoffel symbols and invariant divergence formulas.
Theorem 3.1 (Derivative of Metric Determinant): Let $g = \det(g_{ij})$. Then:
or equivalently:
Proof: Expanding determinant $g$ along its $i$-th row:
Differentiating with respect to $g_{ij}$ yields $\frac{\partial g}{\partial g_{ij}} = G^{ij} = g \, g^{ij}$. By the multivariable chain rule:
Similarly, differentiating the relation $g^{ij} g_{jk} = \delta_k^i$:
§3.3 Tangent Vectors, Reciprocal Bases ($\mathbf{e}_i, \mathbf{e}^i$) & Coordinate Frames
1. Natural Tangent Basis Vectors $\mathbf{e}_i$
Let $\mathbf{r}(x^1, \dots, x^n)$ be the position vector in an embedding space. The natural coordinate basis vectors tangent to the coordinate curves are:
The infinitesimal displacement vector is:
Computing the squared arc length:
Comparing directly with $ds^2 = g_{ij} dx^i dx^j$, we obtain the profound geometric identity:
The components of the metric tensor are the scalar products of the natural tangent basis vectors!
2. Reciprocal Dual Basis Vectors $\mathbf{e}^i$
Definition 3.3 (Dual Reciprocal Basis): The reciprocal basis vectors $\mathbf{e}^1, \mathbf{e}^2, \dots, \mathbf{e}^n$ are the unique vectors satisfying the biorthogonality condition:
In terms of the metric tensor and conjugate metric:
Computing their mutual inner products:
Thus, the conjugate metric components are the scalar products of the reciprocal dual basis vectors:
§3.4 Raising and Lowering Indices: Associated Vectors & Tensors
1. The Musical Isomorphisms ($\flat$ and $\sharp$)
The metric tensor establishes a canonical linear isomorphism between the tangent space $T_p M$ (contravariant vectors) and cotangent space $T_p^* M$ (covariant covectors). In differential geometry, these are called the musical isomorphisms:
- Flat ($\flat$, Index Lowering): Converts a contravariant vector into a covariant covector.
- Sharp ($\sharp$, Index Raising): Converts a covariant covector into a contravariant vector.
2. Algebraic Index Manipulation
For any contravariant vector $A^i$:
Conversely, for any covariant vector $A_i$:
Consistency verification using the inverse relationship:
The operations of raising and lowering indices are exact mutual inverses!
Raising and Lowering in Higher-Rank Tensors:
The metric tensor raises and lowers individual indices independently:
- Lowering the first index of $T^{ij}$:
- Lowering both indices of $T^{ij}$:
- Mixed associated tensor from covariant tensor $T_{ij}$:
(Note: If $T_{ij}$ is not symmetric, the position of the dot placeholder matters: $T^i_{\; j} \ne T_j^{\; i}$).
§3.5 Lengths, Angles & Orthogonal Coordinate Systems
1. Vector Magnitude and Angle between Vectors
In a Riemannian space with metric $g_{ij}$:
1. Magnitude (Norm) of a Vector:
The length $\|A\|$ of a contravariant vector $A^i$ is defined by:
For a covariant vector $B_i$:
2. Angle between Two Vectors:
The angle $\theta$ between two non-zero contravariant vectors $A^i$ and $B^i$ is:
By the Cauchy-Schwarz inequality for positive-definite metrics, $-1 \le \cos\theta \le 1$.
3. Orthogonality Condition:
Two vectors $A^i$ and $B^i$ are orthogonal if and only if their Riemannian inner product vanishes:
2. Orthogonal Curvilinear Coordinate Systems
A coordinate system is orthogonal if the coordinate lines intersect at right angles everywhere. In an orthogonal coordinate system:
The metric tensor and its conjugate are purely diagonal:
The metric determinant is simply the product of diagonal entries:
where $h_i = \sqrt{g_{ii}}$ are the Lamé scale factors.
Rigorous Tiered Solved Examination Problems
Step-by-step unskipped derivations, complete proofs, and verification across Foundational, Advanced, and Honors tiers.
- Cylindrical coordinates $(\bar{x}^1, \bar{x}^2, \bar{x}^3) = (r, \theta, z)$ are defined by:
Compute the covariant metric tensor components $g_{ij}$, the conjugate metric components $g^{ij}$, and the determinant $g$.
- Spherical polar coordinates $(\bar{x}^1, \bar{x}^2, \bar{x}^3) = (r, \theta, \phi)$ have metric:
Find the matrices $(g_{ij})$ and $(g^{ij})$.
- Given a contravariant vector with components $A^i = (1, 2, 3)$ in spherical coordinates at a point where $r = 2, \theta = \pi/2$, compute its associated covariant components $A_i$ and evaluate its invariant length $\|A\| = \sqrt{A^i A_i}$.
1. Cylindrical Coordinates Metric Tensor
In Cartesian coordinates $ds^2 = (dx^1)^2 + (dx^2)^2 + (dx^3)^2$. Differentiating the coordinate relations:
Squaring and adding:
Thus:
Matching with $ds^2 = g_{ij} d\bar{x}^i d\bar{x}^j$ where $\bar{x} = (r, \theta, z)$:
The metric determinant is:
The conjugate metric tensor is the matrix inverse:
2. Spherical Polar Coordinates Metric Matrices
From $ds^2 = dr^2 + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2$ with $\bar{x} = (r, \theta, \phi)$:
Determinant:
The conjugate metric is:
3. Lowering Indices and Computing Invariant Length
At $r = 2, \theta = \pi/2$:
All off-diagonal entries are 0. The contravariant vector components are $A^1 = 1, A^2 = 2, A^3 = 3$. Lowering indices via $A_i = g_{ij} A^j$:
- $A_1 = g_{11} A^1 = 1 \times 1 = 1$
- $A_2 = g_{22} A^2 = 4 \times 2 = 8$
- $A_3 = g_{33} A^3 = 4 \times 3 = 12$
Thus, the covariant components are:
Now compute the invariant scalar magnitude:
Notice that calculating via $\|A\|^2 = g_{ij} A^i A^j = 1(1)^2 + 4(2)^2 + 4(3)^2 = 1 + 16 + 36 = 53$ yields the exact same invariant. $\blacksquare$
Let $V_n$ be a Riemannian space with metric $g_{ij}$.
- Find the unit tangent vectors $\mathbf{u}_{(i)}$ along the $i$-th coordinate curves (where only $x^i$ varies while all other coordinates are held constant).
- Compute the cosine of the Riemannian angle $\theta_{ij}$ between the $i$-th and $j$-th coordinate curves.
- Prove that the coordinate curves are mutually orthogonal at a point if and only if all off-diagonal metric components vanish at that point:
- In an oblique coordinate system in $\mathbb{R}^2$ with axes meeting at angle $\omega$, compute the metric tensor $g_{ij}$ and its conjugate $g^{ij}$.
1. Tangent Vectors along Coordinate Curves
Along the $i$-th coordinate curve, all coordinates except $x^i$ are held fixed. Thus, the differential displacement vector has components:
The Riemannian arc length along this curve is:
The unit tangent vector $u_{(i)}^k$ along the $i$-th coordinate curve is:
2. Cosine of the Angle $\theta_{ij}$ between Coordinate Curves
The Riemannian angle $\theta_{ij}$ between the $i$-th coordinate curve (direction $u_{(i)}$) and $j$-th coordinate curve (direction $u_{(j)}$) is given by:
Substitute $u_{(i)}^k = \frac{\delta_i^k}{\sqrt{g_{ii}}}$ and $u_{(j)}^l = \frac{\delta_j^l}{\sqrt{g_{jj}}}$:
3. Proof of Mutual Orthogonality Criterion
The coordinate curves $x^i$ and $x^j$ ($i \ne j$) are orthogonal if and only if $\theta_{ij} = \pi/2$, which means:
Since $g_{ii} > 0$ and $g_{jj} > 0$ in a positive-definite Riemannian space:
Therefore, the coordinate curves are mutually orthogonal if and only if the metric tensor is purely diagonal ($g_{ij} = 0$ for all $i \ne j$). $\blacksquare$
4. Oblique Coordinates in $\mathbb{R}^2$ with Angle $\omega$
In Cartesian coordinates, let the $x^1$-axis lie along the $x$-axis and the $x^2$-axis be tilted at angle $\omega$:
Differentiating:
Computing the metric components:
- $g_{11} = \mathbf{e}_1 \cdot \mathbf{e}_1 = 1$
- $g_{12} = \mathbf{e}_1 \cdot \mathbf{e}_2 = \cos\omega$
- $g_{22} = \mathbf{e}_2 \cdot \mathbf{e}_2 = \cos^2\omega + \sin^2\omega = 1$
Thus, the metric tensor is:
Determinant:
The conjugate metric tensor is:
Consider the Poincaré upper half-plane model of 2D hyperbolic geometry:
equipped with the Riemannian metric:
where $x^1 = x, x^2 = y$.
- Write down the metric matrix $(g_{ij})$, compute its determinant $g$, and find the conjugate metric $(g^{ij})$.
- Express the hyperbolic metric in complex coordinate notation $z = x + i y \in \mathbb{C}$ with $y = \text{Im}(z) = \frac{z - \bar{z}}{2i}$.
- Consider a general Möbius transformation:
Prove rigorously that the metric $ds^2$ is strictly invariant under all such transformations ($d\tilde{s}^2 = ds^2$).
- Compute the hyperbolic distance between two points $(0, y_1)$ and $(0, y_2)$ along the vertical $y$-axis.
1. Metric Components and Conjugate Metric of $\mathbb{H}^2$
From $ds^2 = \frac{1}{(x^2)^2} (dx^1)^2 + \frac{1}{(x^2)^2} (dx^2)^2$:
Determinant:
Conjugate metric tensor:
2. Complex Coordinate Representation
Let $z = x + i y$. Then $dz = dx + i dy$ and $d\bar{z} = dx - i dy$. The Euclidean numerator is:
The imaginary part is $\text{Im}(z) = y$. Therefore, the hyperbolic metric is expressed compactly as:
3. Invariance under Möbius Transformations $PSL(2, \mathbb{R})$
Let $w = \frac{a z + b}{c z + d}$ with $a, b, c, d \in \mathbb{R}$ and $ad - bc = 1$.
Step A: Differential $dw$
Since $ad - bc = 1$:
Step B: Imaginary Part $\text{Im}(w)$
Step C: Ratio for Transformed Metric
Now assemble the transformed hyperbolic metric $d\tilde{s}^2$:
The conformal denominators $|c z + d|^4$ cancel out identically! Thus, $ds^2$ is strictly invariant under the entire Möbius group $PSL(2, \mathbb{R})$ of hyperbolic isometries. $\blacksquare$
4. Hyperbolic Distance along the Vertical Line $x = 0$
Along the line $x = 0$, $dx = 0$, so $ds = \frac{dy}{y}$. Integrating from $y_1$ to $y_2$ (assuming $y_2 > y_1 > 0$):
Notice that as $y_1 \to 0$ (approaching the boundary real axis), the hyperbolic distance diverges to $\infty$:
The boundary of the upper half-plane is infinitely far away in hyperbolic geometry! $\blacksquare$