What Is the Invariance of Differential Forms?

Haifei included in Mathematical Methods in Physics

2025-03-21 1936 words 10 minutes views

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What Is the Invariance of Differential Forms?

According to textbooks, the invariance of differential forms means that, under a change of variables, the form of a differential equation remains unchanged. That is:

If, when \(\mathrm{d}y\) in \(y\) denotes the function \(y=f(x)\) and \(x\) in \(\mathrm{d}x\) denotes the function \(\phi(x)=x\), \(\mathrm{d}y=a\mathrm{d}x\) holds, that is, \(\mathrm{d}f=a\mathrm{d}\phi\) holds,
then, when \(y\) in \(\mathrm{d}y\) denotes the function \(y=g(t)\) and \(x\) in \(\mathrm{d}x\) denotes the function \(x=\varphi(t)\), \(\mathrm{d}y=a\mathrm{d}x\) likewise holds, that is, \(\mathrm{d}g=a\mathrm{d}\varphi\) holds.
\(\mathrm{d}f=a\mathrm{d}\phi\) and \(\mathrm{d}g=a\mathrm{d}\varphi\) are two different equations, but both can be written as \(\mathrm{d}y=a\mathrm{d}x\). This is the invariance of differential forms.

But is it really necessary to make things this convoluted?

In fact, from a higher-level perspective, the so-called invariance of differential forms is self-evident.

Consider a differentiable function \(y: M \rightarrow \mathbb{R}\) on the one-dimensional manifold \(M\). Its differential is a cotangent vector field on the manifold, or a 1-form: \(\mathrm{d}y: \mathcal{T}^1_0(M) \rightarrow \mathcal{F}(M)\), where \(\mathcal{T}_0^1(M)\) denotes the set of all differentiable tangent vector fields on \(M\), and \(\mathcal{F}\) denotes the set of all differentiable functions on \(M\).

We assign two coordinate systems \(x: M \rightarrow \mathbb{R}^1\) and \(t: M\rightarrow \mathbb{R}^1\) to \(M\), respectively. Coordinate systems themselves are also functions on \(M\).

The validity of \(\mathrm{d} y = a \mathrm{d} x\) (where \(a\) is a function on \(M\)) does not depend on any coordinate system. This is because \(\mathrm{d}y\) depends only on \(y\), and \(\mathrm{d}x\) depends only on \(x\). In other words, whether you regard \(y\) and \(x\) as functions of \(x\) or as functions of \(t\) does not affect the validity of the equation \(\mathrm{d} y = a \mathrm{d} x\).

In differential geometry, \(y\) and \(x\) are both differentiable functions on \(M\), and their existence does not depend on any coordinate system. Therefore, the existence of \(\mathrm{d}y\) and \(\mathrm{d}x\) likewise does not depend on any coordinate system. Like a statue, no matter from which angle you view it, none of its properties are affected.

Thus, we can interpret “the invariance of differential forms” as “\(\mathrm{d}f\) depends only on \(f\) and does not depend on the coordinate system.”

What Is the Difference Between Partial Derivatives and Differentials?

Partial derivatives and differentials are very different. Taking a partial derivative requires specifying a coordinate system; without a coordinate system, one cannot take partial derivatives. When the coordinate system changes, the partial derivatives change accordingly. More critically: even if \(x_1\) does not change, as long as \(x_2,\cdots,x_n\) changes, \(\frac{\partial f}{\partial x_1}\) also changes.

Clearly, if we study only partial derivatives, we become lost in coordinate transformations and find it difficult to abstract to a higher level.

Why must we study not only derivatives but also differentials? This is why: differentials do not depend on a coordinate system, or in other words, they are canonical.

For example, a coordinate-independent canonical 1-form can be defined on the cotangent bundle. According to the preceding discussion, its differential is also canonical. The differential of the canonical 1-form is the symplectic form. Therefore, every cotangent bundle automatically becomes a symplectic manifold.

What Is the Use of the Invariance of Differential Forms?

The most direct application is to “prove” the chain rule:

Substituting \(\mathrm{d}x =\frac{\partial x}{\partial t} \mathrm{d}t\) into \(\mathrm{d}y = \frac{\partial y}{\partial x} \mathrm{d}x\) gives
\(\mathrm{d}y = \frac{\partial y}{\partial x} \mathrm{d}x = \frac{\partial y}{\partial x}\frac{\partial x}{\partial t} \mathrm{d}t\)

In this simple proof, we used “substitution.”

In fact, this is a pseudo-proof. In reality, the chain rule came first, and only then came the invariance of differential forms.

Using “substitution,” we can also prove many useful formulas. Here is an example:

Example 1

(Thermodynamics) Given \(\mathrm{d}E=T\mathrm{d}S-p\mathrm{d}V\), that is, \(\left.\left(\frac{\partial E}{\partial S}\right)\right|_V=T,\,\left.\left(\frac{\partial E}{\partial V}\right)\right|_S=-p\), find \(\left.\left(\frac{\partial E}{\partial T}\right)\right|_V,\,\left.\left(\frac{\partial E}{\partial V}\right)\right|_T\)

Solution: Substituting \(\mathrm{d}S=\left.\left(\frac{\partial S}{\partial T}\right)\right|_V\mathrm{d}T+\left.\left(\frac{\partial S}{\partial V}\right)\right|_T\mathrm{d}V\) into \(\mathrm{d}E=T\mathrm{d}S-p\mathrm{d}V\) gives
\(\begin{aligned} \mathrm{d}E&=T\mathrm{d}S-p\mathrm{d}V \\ &=T\left[\left.\left(\frac{\partial S}{\partial T}\right)\right|_V\mathrm{d}T+\left.\left(\frac{\partial S}{\partial V}\right)\right|_T\mathrm{d}V\right]-p\mathrm{d}V \\ &=T\left.\left(\frac{\partial S}{\partial T}\right)\right|_V\mathrm{d}T+\left[\left.T\left(\frac{\partial S}{\partial V}\right)\right|_T-p\right]\mathrm{d}V \end{aligned}\)
Therefore, \(\left.\left(\frac{\partial E}{\partial T}\right)\right|_V=T\left.\left(\frac{\partial S}{\partial T}\right)\right|_V,\) \(\left.\left(\frac{\partial E}{\partial V}\right)\right|_T=T\left.\left(\frac{\partial S}{\partial V}\right)\right|_T-p\)

Not only can we prove formulas through “substitution”; we can also prove formulas through simple “transposition.” Here are some examples:

Example 2

Solution: Rearranging \(\mathrm{d}E=T\mathrm{d}S-p\mathrm{d}V \) gives \(\mathrm{d}S=\frac{1}{T}\mathrm{d}E-\frac{p}{T}\mathrm{d}V\), so \(\left.\left(\frac{\partial S}{\partial V}\right)\right|_E=\frac{p}{T}\)

Example 3

Given that the partial derivatives of \(z=z(x,y)\) and those of its inverse function exist, prove: \(\frac{\partial z}{\partial x}\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}=-1\)

Proof: Let \(\mathrm{d}z=a\mathrm{d}x+b\mathrm{d}y\), and let \(a\) and \(b\) be nonzero (because the partial derivatives of the inverse function exist). Rearranging gives \(\mathrm{d}x=\frac{1}{a}\mathrm{d}z-\frac{b}{a}\mathrm{d}y\) and \(\mathrm{d}y=\frac{1}{b}\mathrm{d}z-\frac{a}{b}\mathrm{d}x\), so \(\frac{\partial z}{\partial x}\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}=a\cdot\left(-\frac{b}{a}\right)\cdot\left(\frac{1}{b}\right)=-1\)

As can be seen, in these examples, we treat differential forms such as \(\mathrm{d}E\), \(\mathrm{d}S\), and \(\mathrm{d}V\) as ordinary variables for rearrangement and substitution. This is natural, because they are all 1-forms, whose definitions do not depend on any coordinate system.

In proving the above formulas, we have in fact unknowingly used the inverse function theorem and the implicit function theorem.

Example 4

“Prove” the inverse function theorem:

If \(\mathbf{y} : \mathbb{R}^n\rightarrow \mathbb{R}^n\) is differentiable and its derivative (Jacobian matrix) is invertible, then \(\frac{\partial \mathbf{x}}{\partial \mathbf{y}}=\left(\frac{\partial \mathbf{y}}{\partial \mathbf{x}}\right)^{-1}\).

Proof: Denote the Jacobian matrix by
\(\frac{\partial \mathbf{y}}{\partial \mathbf{x}}= \begin{bmatrix} \frac{\partial y_1}{\partial x_1}&\frac{\partial y_1}{\partial x_2}&\cdots&\frac{\partial y_1}{\partial x_n}\\ \frac{\partial y_2}{\partial x_1}&\frac{\partial y_2}{\partial x_2}&\cdots&\frac{\partial y_2}{\partial x_n}\\ \vdots&\vdots&\ddots&\vdots\\ \frac{\partial y_n}{\partial x_1}&\frac{\partial y_n}{\partial x_2}&\cdots&\frac{\partial y_n}{\partial x_n} \end{bmatrix}\)
Then \(\mathrm{d}\mathbf{y} = \frac{\partial \mathbf{y}}{\partial \mathbf{x}}\mathrm{d} \mathbf{x}\)
that is,
\(\begin{bmatrix} \mathrm{d}y_1 \\ \mathrm{d}y_2 \\ \vdots \\ \mathrm{d}y_n \end{bmatrix} = \begin{bmatrix} \frac{\partial y_1}{\partial x_1}&\frac{\partial y_1}{\partial x_2}&\cdots&\frac{\partial y_1}{\partial x_n}\\ \frac{\partial y_2}{\partial x_1}&\frac{\partial y_2}{\partial x_2}&\cdots&\frac{\partial y_2}{\partial x_n}\\ \vdots&\vdots&\ddots&\vdots\\ \frac{\partial y_n}{\partial x_1}&\frac{\partial y_n}{\partial x_2}&\cdots&\frac{\partial y_n}{\partial x_n} \end{bmatrix} \begin{bmatrix} \mathrm{d}x_1 \\ \mathrm{d}x_2 \\ \vdots \\ \mathrm{d}x_n \end{bmatrix}\)
Solving gives \(\begin{bmatrix} \mathrm{d}x_1 \\ \mathrm{d}x_2 \\ \vdots \\ \mathrm{d}x_n \end{bmatrix} = \begin{bmatrix} \frac{\partial y_1}{\partial x_1}&\frac{\partial y_1}{\partial x_2}&\cdots&\frac{\partial y_1}{\partial x_n}\\ \frac{\partial y_2}{\partial x_1}&\frac{\partial y_2}{\partial x_2}&\cdots&\frac{\partial y_2}{\partial x_n}\\ \vdots&\vdots&\ddots&\vdots\\ \frac{\partial y_n}{\partial x_1}&\frac{\partial y_n}{\partial x_2}&\cdots&\frac{\partial y_n}{\partial x_n} \end{bmatrix}^{-1} \begin{bmatrix} \mathrm{d}y_1 \\ \mathrm{d}y_2 \\ \vdots \\ \mathrm{d}y_n \end{bmatrix}\)
that is, \(\mathrm{d}\mathbf{x} = \left(\frac{\partial \mathbf{y}}{\partial \mathbf{x}}\right)^{-1}\mathrm{d} \mathbf{y}\)
Therefore, \(\frac{\partial \mathbf{x}}{\partial \mathbf{y}}=\left(\frac{\partial \mathbf{y}}{\partial \mathbf{x}}\right)^{-1}\)

Example 5

“Prove” the implicit function theorem:

Given \(f:\mathbb{R}^m\times\mathbb{R}^n\rightarrow\mathbb{R^n}\), \(\mathbf f(x_1,x_2,\cdots,x_m,y_1,y_2,\cdots,y_n)=\bm{0}\), if \(\frac{\partial\mathbf{f}}{\partial\mathbf{y}}\) is invertible, then \(\frac{\partial \mathbf y}{\partial \mathbf x}=-\left(\frac{\partial \mathbf f}{\partial \mathbf y}\right)^{-1}\frac{\partial \mathbf f}{\partial \mathbf x}\)

Proof: From \(\mathbf{f}=\mathbf{0}\), we obtain \(\mathrm{d} \mathbf{f} = \mathbf{0}\), that is,
\(\begin{bmatrix} \mathrm{d}f_1 \\ \vdots \\ \mathrm{d}f_m \end{bmatrix} = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_m} & \frac{\partial f_1}{\partial y_1} & \cdots & \frac{\partial f_1}{\partial y_n} \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ \frac{\partial f_{m}}{\partial x_1} & \cdots & \frac{\partial f_{m}}{\partial x_m} & \frac{\partial f_{m}}{\partial y_1} & \cdots & \frac{\partial f_{m}}{\partial y_n} \end{bmatrix} \begin{bmatrix} \mathrm{d} x_1 \\ \vdots \\ \mathrm{d} x_m \\ \mathrm{d}y_1 \\ \vdots \\ \mathrm{d}y_n \end{bmatrix} = \begin{bmatrix} 0 \\ \vdots \\ 0 \end{bmatrix}\)
Rearranging gives:
\(\begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_m} \\ \vdots & \ddots & \vdots \\ \frac{\partial f_{m}}{\partial x_1} & \cdots & \frac{\partial f_{m}}{\partial x_m} \end{bmatrix} \begin{bmatrix} \mathrm{d} x_1 \\ \vdots \\ \mathrm{d} x_m \end{bmatrix} = -\begin{bmatrix} \frac{\partial f_1}{\partial y_1} & \cdots & \frac{\partial f_1}{\partial y_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial f_{m}}{\partial y_1} & \cdots & \frac{\partial f_{m}}{\partial y_n} \end{bmatrix} \begin{bmatrix} \mathrm{d}y_1 \\ \vdots \\ \mathrm{d}y_n \end{bmatrix}\)
Solving gives:
\(\begin{bmatrix} \mathrm{d}y_1 \\ \vdots \\ \mathrm{d}y_n \end{bmatrix} = -\begin{bmatrix} \frac{\partial f_1}{\partial y_1} & \cdots & \frac{\partial f_1}{\partial y_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial f_{m}}{\partial y_1} & \cdots & \frac{\partial f_{m}}{\partial y_n} \end{bmatrix}^{-1} \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_m} \\ \vdots & \ddots & \vdots \\ \frac{\partial f_{m}}{\partial x_1} & \cdots & \frac{\partial f_{m}}{\partial x_m} \end{bmatrix} \begin{bmatrix} \mathrm{d} x_1 \\ \vdots \\ \mathrm{d} x_m \end{bmatrix}\)
that is, \(\mathrm{d}\mathbf{y}=\left[-\left(\frac{\partial \mathbf f}{\partial \mathbf y}\right)^{-1}\frac{\partial \mathbf f}{\partial \mathbf x}\right]\mathrm{d}\mathbf{x}\)

Thus, to some extent, both the inverse function theorem and the implicit function theorem are almost self-evident. Indeed, in textbooks, they are both proved using the chain rule, and the chain rule itself is the most direct manifestation of “the invariance of differential forms.”

The inverse function theorem and the implicit function theorem are the most widely applicable tools. Exercises 1–3 can all be proved using the inverse function theorem and the implicit function theorem.

What Are Higher-Order Differentials?

Higher-order differentials are meaningless, because taking the exterior differential of any function twice yields zero: \(\mathrm{d}\mathrm{d}=0\).

Although higher-order (exterior) differentials have little meaning, “higher-degree differential forms” are quite meaningful. In differential geometry, “differential form” is a technical term. We have first-degree differential forms, second-degree differential forms, …, and nth-degree differential forms, or simply n-forms.

“Higher-degree differential forms” still possess the “invariance” described above. For example, the definition of \(\mathrm{d}f \wedge \mathrm{d}g\) depends only on the definitions of \(f\) and \(g\), and does not depend on any coordinate system.

For example, in thermodynamics, we have:

\[\mathrm{d} T \wedge \mathrm{d}S = \mathrm{d} p \wedge \mathrm{d} V\]

Proof: \(\mathrm{d} T \wedge \mathrm{d}S - \mathrm{d} p \wedge \mathrm{d} V= \mathrm{d}(T\mathrm{d}S - p \mathrm{d}V) = \mathrm{d}\mathrm{d}E = 0\)

It holds independently of any local coordinate system. Using this, we can prove the Maxwell relations as follows:

\[\begin{aligned} \mathrm{d} T \wedge \mathrm{d}S &=\left[\left(\frac{\partial T}{\partial V}\right)_S \mathrm{d}V + \left(\frac{\partial T}{\partial S}\right)_V \mathrm{d}S\right] \wedge \mathrm{d}S \\ &= \left(\frac{\partial T}{\partial V}\right)_S \mathrm{d}V \wedge \mathrm{d}S \end{aligned}\]\[\begin{aligned} \mathrm{d} p \wedge \mathrm{d}V &=\left[\left(\frac{\partial p}{\partial V}\right)_S \mathrm{d}V + \left(\frac{\partial p}{\partial S}\right)_V \mathrm{d}S\right] \wedge \mathrm{d}V \\ &= \left(\frac{\partial p}{\partial S}\right)_V \mathrm{d}S \wedge \mathrm{d}V \\ &= - \left(\frac{\partial p}{\partial S}\right)_V \mathrm{d}V \wedge \mathrm{d}S \end{aligned}\]

Therefore, \(\begin{aligned} \left(\frac{\partial T}{\partial V}\right)_S = - \left(\frac{\partial p}{\partial S}\right)_V \end{aligned}\)

The other Maxwell relations can also be obtained by this method.

Thus, not only do “first-degree differential forms,” or 1-forms, possess invariance, but “nth-degree differential forms,” or n-forms, possess invariance as well. And they are also very useful.

Analysis vs. Geometry

The invariance of differential forms is based on the chain rule, the implicit function theorem, and the inverse function theorem.

In elementary calculus (mathematical analysis), the invariance of differential forms is a manifestation of these theorems. In other words, when applying the invariance of differential forms, we are in fact using these theorems.

From the perspective of differential geometry, however, the invariance of differential forms is self-evident. Of course, differential geometry is built on elementary calculus.

Returning to the initial example:

If, when \(y\) in \(\mathrm{d}y\) denotes the function \(y=f(x)\) and \(x\) in \(\mathrm{d}x\) denotes the function \(\phi(x)=x\), \(\mathrm{d}y=a\mathrm{d}x\) holds, that is, \(\mathrm{d}f=a\mathrm{d}\phi\) holds,
then, when \(y\) in \(\mathrm{d}y\) denotes the function \(y=g(t)\) and \(x\) in \(\mathrm{d}x\) denotes the function \(x=\varphi(t)\), \(\mathrm{d}y=a\mathrm{d}x\) likewise holds, that is, \(\mathrm{d}g=a\mathrm{d}\varphi\) holds.
\(\mathrm{d}f=a\mathrm{d}\phi\) and \(\mathrm{d}g=a\mathrm{d}\varphi\) are two different equations, but both can be written as \(\mathrm{d}y=a\mathrm{d}x\). This is the invariance of differential forms.

From the perspective of analysis, the functions \(f\) and \(g\) above are different functions, and the functions \(\phi\) and \(\varphi\) are also different functions. Meanwhile, \(y,\ x,\ t\) are merely variables, not the functions themselves.

From the perspective of geometry, however, \(y\) is the function \(y\); \(x\) is the function \(x\); and \(t\) is the function \(t\). They are all functions on the manifold.

Consider another example (the inverse function theorem):

Rearranging \(\mathrm{d}y = a\mathrm{d}x\) gives \(\mathrm{d}x = \frac{1}{a}\mathrm{d}y\)

From the perspective of analysis, this operation applies the inverse function theorem rather than being a simple rearrangement. This is because the function changes from \(y=f(x)\) to \(x=g(y)\), and \(f\) and \(g\) are two different functions.

But from the perspective of geometry, this really is an entirely ordinary rearrangement.

Consider another example (the implicit function theorem):

Rearranging \(\mathrm{d}z = a\mathrm{d}x + b \mathrm{d}y\) gives \(\mathrm{d}x = \frac{1}{a}\mathrm{d}z - \frac{b}{a}\mathrm{d}y\)

From the perspective of analysis, this actually applies the implicit function theorem rather than being a simple rearrangement. This is because the function changes from \(z=f(x,y)\) to \(x=g(y,z)\), and \(f\) and \(g\) are two different functions.

But from the perspective of geometry, this really is an entirely ordinary rearrangement.