The pullback of a k-form

Section 4.7 The pullback of a $k$ -form

In Section 2.4 we defined the pullback of a one-form. We now generalize this concept to

k

-forms. We first do it using an axiomatic approach as for one-forms, and relate the result to the concept of Jacobian. We also introduce a more direct definition of pullback using the algebraic approach to basic

k

-forms introduced in Subsection 4.1.1, and show that it is consistent with our axiomatic approach..

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Objectives

You should be able to:

Determine the pullback of a $k$ -form.
Show that the pullback of a $2$ -form in $R^{2}$ and a $3$ -form in $R^{3}$ is given by the Jacobian determinant.

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Subsection 4.7.1 The pullback of a $k$ -form: an axiomatic approach

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In Section 2.4 we defined the pullback of a one-form. We used an “axiomatic” approach: we first defined the properties that we wanted the pullback to satisfy, and showed that there is a unique construction that satisfies these properties. The properties that we specified were, in words:

The pullback of a sum of one-forms is the sum of the pullbacks.
The pullback of a product of a zero-form and a one-form is the product of the pullbacks.
The pullback of the exterior derivative of a zero-form is the exterior derivative of the pullback.

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In this section we define the pullback of a

k

-form using a similar approach. All we need to do is generalize the first and second properties to

k

-forms.

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More precisely, let

ϕ : V \to U

be a smooth function, with

V \subseteq R^{m}

and

U \subseteq R^{n}

open subsets. We want the pullback

ϕ^{*}

to satisfy the following properties:

If $ω$ and $η$ are $k$ -forms on $U,$ then
$ϕ^{*} (ω + η) = ϕ^{*} ω + ϕ^{*} η .$
If $ω$ is a $k$ -form and $η$ an $ℓ$ -form on $U,$ then
$ϕ^{*} (ω \land η) = (ϕ^{*} ω) \land (ϕ^{*} η) .$
If $f$ is a zero-form (a function) on $U,$ then
$ϕ^{*} (d f) = d (ϕ^{*} f) .$

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The third property is unchanged, and the first two are naturally generalized to

k

-forms.

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Imposing these properties gives us a unique definition for the pullback of a

k

-form.

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Lemma 4.7.1. The pullback of a $k$ -form.

Let $ϕ : V \to U$ be a smooth function, with $V \subseteq R^{m}$ and $U \subseteq R^{n}$ open subsets. We write $t = (t_{1}, \dots, t_{m}) \in V,$ and $ϕ (t) = (x_{1} (t), \dots, x_{n} (t)) .$ Let $ω$ be a $k$ -form on $U,$ with

ω = \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} f_{i_{1} \dots i_{k}} d x_{i_{1}} \land \dots \land d x_{i_{k}},

for some functions $f_{i_{1} \dots i_{k}} : U \to R .$ Then the pullback $ϕ^{*} ω$ is a $k$ -form on $V$ given by:

\begin{aligned} ϕ^{*} ω = \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} f_{i_{1} \dots i_{k}} (ϕ (t)) & (\frac{\partial x_{i_{1}}}{\partial t_{1}} d t_{1} + \dots + \frac{\partial x_{i_{1}}}{\partial t_{m}} d t_{m}) \\ \land \dots \land (\frac{\partial x_{i_{k}}}{\partial t_{1}} d t_{1} + \dots + \frac{\partial x_{i_{k}}}{\partial t_{m}} d t_{m}) . \end{aligned}

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Proof.

Start with a basic $k$ -form

d x_{i_{1}} \land \dots \land d x_{i_{k}} .

Since we impose (Property 2) that $ϕ^{*} (ω \land η) = (ϕ^{*} ω) \land ϕ^{*} η),$ we know that

ϕ^{*} (d x_{i_{1}} \land \dots \land d x_{i_{k}}) = ϕ^{*} (d x_{i_{1}}) \land \dots \land ϕ^{*} (d x_{i_{k}}) . .

But we already calculated how basic one-forms transform in Lemma 2.4.4; they transform as differentials would. This calculation followed from Property 3, which we still impose here, so it is still valid. This gives us the pullback of basic $k$ -forms.

Then we use Property 2 to extended it to a function times a basic $k$ -form, and then Property 1 to extend it to linear combination of such terms, to get the final result for the pullback of a generic $k$ -form.

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This formula certainly looks ugly, but it is really not that bad. Concretely, all you need to do is compose the component functions with

ϕ,

and transform the basic one-forms one by one in the wedge product as in Lemma 2.4.4, i.e. they transform as you would expect differentials transform. That's it! This will certainly be clearer with examples.

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Example 4.7.2. The pullback of a two-form.

Consider the following two-form on $R^{3} :$

ω = x y d y \land d z + (x z + y) d z \land d x + d x \land d y,

and the smooth function $ϕ : R^{2} \to R^{3}$ given by

ϕ (u, v) = (u v, u^{2}, v^{2}) = (x (u, v), y (u, v), z (u, v)) .

To calculate $ϕ^{*} ω,$ let us start by calculating the pullback of the basic one-forms. We get:

\begin{aligned} ϕ^{*} (d x) = & \frac{\partial x}{\partial u} d u + \frac{\partial x}{\partial v} d v = v d u + u d v, \\ ϕ^{*} (d y) = & \frac{\partial y}{\partial u} d u + \frac{\partial y}{\partial v} d v = 2 u d u, \\ ϕ^{*} (d z) = & \frac{\partial z}{\partial u} d u + \frac{\partial z}{\partial v} d v = 2 v d v . \end{aligned}

Putting this together, we get:

\begin{aligned} ϕ^{*} ω = & (u v) (u^{2}) ϕ^{*} (d y) \land ϕ^{*} (d z) + ((u v) (v^{2}) + u^{2}) ϕ^{*} (d z) \land ϕ^{*} (d x) + ϕ^{*} (d x) \land ϕ^{*} (d y) \\ = & u^{3} v (2 u d u) \land (2 v d v) + (u v^{3} + u^{2}) (2 v d v) \land (v d u + u d v) + (v d u + u d v) \land (2 u d u) \\ = & 4 u^{4} v^{2} d u \land d v + 2 u v^{2} (v^{3} + u) d v \land d u + 2 u^{2} d v \land d u \\ = & (4 u^{4} v^{2} - 2 u v^{5} - 2 u^{2} v^{2} - 2 u^{2}) d u \land d v . \end{aligned}

Note that we used the fact that $d u \land d u = d v \land d v = 0,$ and $d v \land d u = - d u \land d v .$

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Example 4.7.3. The pullback of a three-form.

Consider the following three-form on $R^{3} :$

ω = e^{x + y + z} d x \land d y \land d z,

and the smooth function $ϕ : R_{> 0}^{3} \to R^{3}$ given by

ϕ (u, v, w) = (\ln (u v), \ln (v w), \ln (w u)) .

The pullback of the basic one-forms is:

\begin{aligned} ϕ^{*} (d x) = & \frac{1}{u} d u + \frac{1}{v} d v, \\ ϕ^{*} (d y) = & \frac{1}{v} d v + \frac{1}{w} d w, \\ ϕ^{*} (d z) = & \frac{1}{w} d w + \frac{1}{u} d u . \end{aligned}

We get:

\begin{aligned} ϕ^{*} ω = & e^{\ln (u v) + \ln (v w) + \ln (w u)} (\frac{1}{u} d u + \frac{1}{v} d v) \land (\frac{1}{v} d v + \frac{1}{w} d w) \land (\frac{1}{w} d w + \frac{1}{u} d u) \\ = & u^{2} v^{2} w^{2} (\frac{1}{u v w} d u \land d v \land d w + \frac{1}{u v w} d v \land d w \land d u) \\ = & 2 u v w d u \land d v \land d w, \end{aligned}

where we used the fact that the basic three-forms vanish whenever one of the factor is repeated, and $d v \land d w \land d u = d u \land d v \land d w$ since we need to exchange two basic one-forms twice to relate the two basic three-forms.

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Good, so we are now experts at computing pullbacks! Calculating the pullback of a

k

-form is not more difficult than calculating the pullback of a one-form, but the calculation may be longer, and you need to use the anti-symmetry properties of basic

k

-forms in Lemma 4.1.6 to simplify the result at the end of your calculation.

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Property 3 in our axiomatic definition states that the pullback commutes with the exterior derivative for zero-form. It turns out that this property, which is very important, holds in general for

k

-forms. Let us prove that.

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Lemma 4.7.4. The pullback commutes with the exterior derivative.

Let $ϕ : V \to U$ be a smooth function, with $V \subseteq R^{m}$ and $U \subseteq R^{n}$ open subsets. Let $ω$ be a $k$ -form on $U .$ Then:

ϕ^{*} (d ω) = d (ϕ^{*} ω) .

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Proof.

Recall the definition of the exterior derivative Definition 4.3.1. Let $ω = \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} f_{i_{1} \dots i_{k}} d x_{i_{1}} \land \dots \land d x_{i_{k}} .$ Then

d ω = \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} d (f_{i_{1} \dots i_{k}}) \land d x_{i_{1}} \land \dots \land d x_{i_{k}} .

Using Properties 1 and 2, we can write

ϕ^{*} (d ω) = \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} ϕ^{*} (d (f_{i_{1} \dots i_{k}})) \land ϕ^{*} (d x_{i_{1}}) \land \dots \land ϕ^{*} (d x_{i_{k}}) .

Now we can use Property 3, which states that

ϕ^{*} (d (f_{i_{1} \dots i_{k}})) = d (ϕ^{*} f_{i_{1} \dots i_{k}}),

since the $f_{i_{1} \dots i_{k}}$ are just functions (zero-forms). Thus we have:

\begin{aligned} ϕ^{*} (d ω) = & \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} d (ϕ^{*} f_{i_{1} \dots i_{k}}) \land ϕ^{*} (d x_{i_{1}}) \land \dots \land ϕ^{*} (d x_{i_{k}}) \\ = & d (\sum_{1 \leq i_{1} < \dots < i_{k} \leq n} (ϕ^{*} f_{i_{1} \dots i_{k}}) ϕ^{*} (d x_{i_{1}}) \land \dots \land ϕ^{*} (d x_{i_{k}})) \\ = & d (ϕ^{*} ω), \end{aligned}

where the last line follows from Properties 1 and 2 again.

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Subsection 4.7.2 The pullback of a top form in $R^{n}$ and the Jacobian determinant

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There is a special case for which the pullback takes a very nice form. This case will play a role shortly in our theory of integration, as it will be related to the transformation formula (the generalization of the substitution formula) for multiple integrals.

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Let us start by recalling the definition of the Jacobian of a smooth function.

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Definition 4.7.5. The Jacobian.

Let $ϕ : V \to U$ be a smooth function with $U, V \subseteq R^{n}$ open subsets. Let us write $t = (t_{1}, \dots, t_{n}) \in V,$ and

ϕ (t) = (x_{1} (t), \dots, x_{n} (t)) .

The Jacobian of $ϕ,$ which we denote by $\frac{\partial (x_{1}, \dots, x_{n})}{\partial (t_{1}, \dots, t_{n})}$ or $J_{ϕ}$ or $D ϕ$ (lots of different notations!), is the $n \times n$ matrix of first partial derivatives:

D ϕ = J_{ϕ} = \frac{\partial (x_{1}, \dots, x_{n})}{\partial (t_{1}, \dots, t_{n})} = (\begin{matrix} \frac{\partial x_{1}}{\partial t_{1}} & \dots & \frac{\partial x_{1}}{\partial t_{n}} \\ ⋮ & ⋱ & ⋮ \\ \frac{\partial x_{n}}{\partial t_{1}} & \dots & \frac{\partial x_{n}}{\partial t_{n}} \end{matrix})

Its determinant is called the Jacobian determinant.

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It turns out that if we pullback an

n

-form with respect to such a

ϕ,

we can write

ϕ^{*} ω

in terms of the Jacobian determinant. First, let us introduce the common name “top form” for

n

-form on open subsets

U \subseteq R^{n} .

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Definition 4.7.6. Top form.

We call an $n$ -form on an open subset $U \subseteq R^{n}$ a top form.

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Such forms are called “top forms” because all forms with

k \geq n

necessarily vanish on

R^{n} .

Going back to the pullback, we get the nice following result when we pullback a top form:

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Lemma 4.7.7. The pullback of a top form in $R^{n}$ in terms of the Jacobian determinant.

Suppose that $ω$ is a top form on $U \subseteq R^{n},$ and $ϕ : V \to U$ a smooth function with $V \subseteq R^{n} .$ As above we write

ϕ (t) = (x_{1} (t), \dots, x_{n} (t))

with $t = (t_{1}, \dots, t_{n}) \in V,$ and

ω = f d x_{1} \land \dots \land d x_{n}

for some function $f : U \to R .$ Then

ϕ^{*} ω = f (ϕ (t)) (det J_{ϕ}) d t_{1} \land \dots \land d t_{n},

where $det J_{ϕ}$ is the Jacobian determinant.

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Proof for $R^{2}$ and $R^{3}$ .

It is not so easy to write a general proof for $R^{n}$ using the computational approach for the pullback that we have used so far. We will be able to write down a general proof easily in the next subsection after having introduced a more algebraic approach to the pullback. For the time being, let us prove the statement explicitly for $R^{2}$ and $R^{3} .$

For $R^{2},$ our function $ϕ$ takes the form

ϕ (t_{1}, t_{2}) = (x_{1} (t_{1}, t_{2}), x_{2} (t_{1}, t_{2})) .

What we need to show is that

ϕ^{*} (d x_{1} \land d x_{2}) = (det J_{ϕ}) d t_{1} \land d t_{2},

where

det J_{ϕ} = det (\begin{matrix} \frac{\partial x_{1}}{\partial t_{1}} & \frac{\partial x_{1}}{\partial t_{2}} \\ \frac{\partial x_{2}}{\partial t_{1}} & \frac{\partial x_{2}}{\partial t_{2}} \end{matrix}) = \frac{\partial x_{1}}{\partial t_{1}} \frac{\partial x_{2}}{\partial t_{2}} - \frac{\partial x_{2}}{\partial t_{1}} \frac{\partial x_{1}}{\partial t_{2}} .

But

\begin{aligned} ϕ^{*} (d x_{1} \land d x_{2}) = & (\frac{\partial x_{1}}{\partial t_{1}} d t_{1} + \frac{\partial x_{1}}{\partial t_{2}} d t_{2}) \land (\frac{\partial x_{2}}{\partial t_{1}} d t_{1} + \frac{\partial x_{2}}{\partial t_{2}} d t_{2}) \\ = & (\frac{\partial x_{1}}{\partial t_{1}} \frac{\partial x_{2}}{\partial t_{2}} - \frac{\partial x_{2}}{\partial t_{1}} \frac{\partial x_{1}}{\partial t_{2}}) d t_{1} \land d t_{2}, \end{aligned}

and the lemma is proved.

The calculation is similar but more involved for $R^{3} .$ We have:

ϕ (t_{1}, t_{2}, t_{3}) = (x_{1} (t_{1}, t_{2}, t_{3}), x_{2} (t_{1}, t_{2}, t_{3}), x_{3} (t_{1}, t_{2}, t_{3})) .

What we need to show is that

ϕ^{*} (d x_{1} \land d x_{2} \land d x_{3}) = (det J_{ϕ}) d t_{1} \land d t_{2} \land d t_{3},

where

\begin{aligned} det J_{ϕ} = & det (\begin{array}{c} \frac{\partial x_{1}}{\partial t_{1}} & \frac{\partial x_{1}}{\partial t_{2}} & \frac{\partial x_{1}}{\partial t_{3}} \\ \frac{\partial x_{2}}{\partial t_{1}} & \frac{\partial x_{2}}{\partial t_{2}} & \frac{\partial x_{2}}{\partial t_{3}} \\ \frac{\partial x_{3}}{\partial t_{1}} & \frac{\partial x_{3}}{\partial t_{2}} & \frac{\partial x_{3}}{\partial t_{3}} \end{array}) \\ = & \frac{\partial x_{1}}{\partial t_{1}} \frac{\partial x_{2}}{\partial t_{2}} \frac{\partial x_{3}}{\partial t_{3}} + \frac{\partial x_{1}}{\partial t_{2}} \frac{\partial x_{2}}{\partial t_{3}} \frac{\partial x_{3}}{\partial t_{1}} + \frac{\partial x_{1}}{\partial t_{3}} \frac{\partial x_{2}}{\partial t_{1}} \frac{\partial x_{3}}{\partial t_{2}} \\ - \frac{\partial x_{1}}{\partial t_{2}} \frac{\partial x_{2}}{\partial t_{1}} \frac{\partial x_{3}}{\partial t_{3}} - \frac{\partial x_{1}}{\partial t_{1}} \frac{\partial x_{2}}{\partial t_{3}} \frac{\partial x_{3}}{\partial t_{2}} - \frac{\partial x_{1}}{\partial t_{3}} \frac{\partial x_{2}}{\partial t_{2}} \frac{\partial x_{3}}{\partial t_{1}} . \end{aligned}

But

\begin{aligned} ϕ^{*} (d x_{1} \land d x_{2} \land d x_{3}) = & (\frac{\partial x_{1}}{\partial t_{1}} d t_{1} + \frac{\partial x_{1}}{\partial t_{2}} d t_{2} + \frac{\partial x_{1}}{\partial t_{3}} d t_{3}) \\ \land (\frac{\partial x_{2}}{\partial t_{1}} d t_{1} + \frac{\partial x_{2}}{\partial t_{2}} d t_{2} + \frac{\partial x_{2}}{\partial t_{3}} d t_{3}) \\ \land (\frac{\partial x_{3}}{\partial t_{1}} d t_{1} + \frac{\partial x_{3}}{\partial t_{2}} d t_{2} + \frac{\partial x_{3}}{\partial t_{3}} d t_{3}) \\ = & (\frac{\partial x_{1}}{\partial t_{1}} \frac{\partial x_{2}}{\partial t_{2}} \frac{\partial x_{3}}{\partial t_{3}} + \frac{\partial x_{1}}{\partial t_{2}} \frac{\partial x_{2}}{\partial t_{3}} \frac{\partial x_{3}}{\partial t_{1}} + \frac{\partial x_{1}}{\partial t_{3}} \frac{\partial x_{2}}{\partial t_{1}} \frac{\partial x_{3}}{\partial t_{2}} \\ - \frac{\partial x_{1}}{\partial t_{2}} \frac{\partial x_{2}}{\partial t_{1}} \frac{\partial x_{3}}{\partial t_{3}} - \frac{\partial x_{1}}{\partial t_{1}} \frac{\partial x_{2}}{\partial t_{3}} \frac{\partial x_{3}}{\partial t_{2}} - \frac{\partial x_{1}}{\partial t_{3}} \frac{\partial x_{2}}{\partial t_{2}} \frac{\partial x_{3}}{\partial t_{1}}) d t_{1} \land d t_{2} \land d t_{3}, \end{aligned}

which completes the proof in $R^{3} .$ Phew, this was painful.

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The appearance of the Jacobian determinant here is quite nice. It will be related to the transformation formula for multiple integrals, when we integrate a top form over a bounded region in

R^{n} .

We note however that we obtain the Jacobian determinant here, not its absolute value (in comparison to what you may have seen in previous calculus classes); this will be related to the fact that our theory of integration is oriented.

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Subsection 4.7.3 The pullback of a $k$ -form: an algebraic approach (optional)

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In this section we introduce a direct definition of the pullback using the algebraic approach to basic

k

-forms introduced in Subsection 4.1.1. We then show that the three fundamental properties that we used to define the pullback are satisfied, thus justifying our original axiomatic approach.

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Recall from Definition 4.1.1 (naturally generalized to

R^{n}

) that a basic one-form

d x_{i}

is a linear map

d x_{i} : R^{n} \to R

which acts as

d x_{i} (u_{1}, \dots, u_{n}) = u_{i},

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i.e. it outputs the

i

'th component of the vector

u \in R^{n} .

As we are now thinking of the basic one-forms

d x_{i}

as linear maps, we can define their pullback by composition, as we originally did for functions in Definition 2.4.1.

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We first define and study the pullback when

ϕ

is a linear map between vector spaces. We will generalize to the case of a smooth function afterwards.

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Definition 4.7.8. The pullback of a basic one-form with respect to a linear map.

Let $d x_{i} : R^{n} \to R$ be a basic one-form, and let $ϕ : R^{m} \to R^{n}$ be a linear map. We define the pullback $ϕ^{*} (d x_{i}) : R^{m} \to R$ by composition:

ϕ^{*} (d x_{i}) = d x_{i} \circ ϕ : R^{m} \to R .

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Let us now give an explicit formula for the pullback of a basic one-form.

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Lemma 4.7.9. An explicit formula for the pullback of a basic one-form.

Let $d x_{i} : R^{n} \to R,$ $i = 1, \dots, n$ be the basic one-forms on $R^{n},$ and $d t_{j} : R^{m} \to R,$ $j = 1, \dots, m$ be the basic one-forms on $R^{m} .$ The linear map $ϕ : R^{m} \to R^{n}$ can be represented by a matrix $A = (a_{i j}) .$ Then we can write

ϕ^{*} (d x_{i}) = \sum_{j = 1}^{m} a_{i j} d t_{j} = a_{i 1} d t_{1} + \dots + a_{i m} d t_{m} .

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Proof.

Write $v = (v_{1}, \dots, v_{m}) \in R^{m} .$ Then

\begin{aligned} ϕ^{*} (d x_{i}) (v) = & d x_{i} (ϕ (v)) \\ = & d x_{i} (a_{11} v_{1} + \dots + a_{1 m} v_{m}, \dots, a_{n 1} v_{1} + \dots a_{n m} v_{m}) \\ = & a_{i 1} v_{1} + \dots + a_{i m} v_{m} \\ = & a_{i 1} d t_{1} (v) + \dots + a_{i m} d t_{m} (v) \\ = & (a_{i 1} d t_{1} + \dots + a_{i m} d t_{m}) (v) . \end{aligned}

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It is then easy to generalize the definition of pullback to the basic

k

-forms.

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Definition 4.7.10. The pullback of a basic $k$ -form with respect to a linear map.

Let $d x_{i_{1}} \land \dots \land d x_{i_{k}} : (R^{n})^{k} \to R$ be a basic $k$ -form, and let $ϕ : R^{m} \to R^{n}$ be a linear map. Let $v^{1}, \dots, v^{k} \in R^{m}$ be vectors. We define the pullback $ϕ^{*} (d x_{i_{1}} \land \dots \land d x_{i_{k}}) : (R^{m})^{k} \to R$ by:

ϕ^{*} (d x_{i_{1}} \land \dots \land d x_{i_{k}}) (v^{1}, \dots, v^{k}) = d x_{i_{1}} \land \dots \land d x_{i_{k}} (ϕ (v^{1}), \dots, ϕ (v^{k})) .

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It follows directly from the definition that the pullback commutes with the wedge product:

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Lemma 4.7.11. The pullback commutes with the wedge product.

Under the setup above,

ϕ^{*} (d x_{i_{1}} \land \dots \land d x_{i_{k}}) = ϕ^{*} (d x_{i_{1}}) \land \dots \land ϕ^{*} (d x_{i_{k}}) .

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Proof.

By definition,

\begin{aligned} ϕ^{*} (d x_{i_{1}}) \land \dots \land ϕ^{*} (d x_{i_{k}}) (v^{1}, \dots, v^{k}) = & d x_{i_{1}} \land \dots \land d x_{i_{k}} (ϕ (v^{1}), \dots, ϕ (v^{k})) \\ = & ϕ^{*} (d x_{i_{1}} \land \dots \land d x_{i_{k}}) (v^{1}, \dots, v^{k}) . \end{aligned}

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As a corollary, we obtain an explicit formula to calculate the pullback of a basic

k

-form.

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Corollary 4.7.12. An explicit formula for the pullback of a basic $k$ -form.

Let $d x_{i_{1}} \land \dots \land d x_{i_{k}} : (R^{n})^{k} \to R$ be a basic $k$ -form, and let $ϕ : R^{m} \to R^{n}$ be a linear map. Let $d t_{j} : R^{m} \to R,$ $j = 1, \dots, m$ be the basic one-forms on $R^{m} .$ The linear map $ϕ : R^{m} \to R^{n}$ can be represented by a matrix $A = (a_{i j}) .$ Then we can write

ϕ^{*} (d x_{i_{1}} \land \dots \land d x_{i_{k}}) = (a_{i_{1} 1} d t_{1} + \dots + a_{i_{1} m} d t_{m}) \land \dots \land (a_{i_{k} 1} d t_{1} + \dots + a_{i_{k} m} d t_{m}) .

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We can also write down an explicit formula in terms of the determinant when we pullback a basic

n

-form from

R^{n}

R^{n} .

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Lemma 4.7.13. The pullback of a basic $n$ -form in $R^{n}$ .

Let $d x_{1} \land \dots \land d x_{n} : (R^{n})^{n} \to R,$ and let $ϕ : R^{n} \to R^{n}$ be a linear map, which can be representated by an $n \times n$ matrix $A = (a_{i j}) .$ Then

ϕ^{*} (d x_{1} \land \dots \land d x_{n}) = (det A) d x_{1} \land \dots \land d x_{n} .

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Proof.

Let $v^{1}, \dots, v^{n} \in R^{n} .$ Then

\begin{aligned} ϕ^{*} (d x_{1} \land \dots \land d x_{n}) (v^{1}, \dots, v^{n}) = & d x_{1} \land \dots \land d x_{n} (ϕ (v^{1}), \dots, ϕ (v^{n})) \\ = & det (\begin{array}{c} A_{11} v_{1}^{1} + \dots + A_{1 n} v_{n}^{1} & \dots & A_{11} v_{1}^{n} + \dots + A_{1 n} v_{n}^{n} \\ ⋮ & ⋱ & ⋮ \\ A_{n 1} v_{1}^{1} + \dots + A_{n n} v_{n}^{1} & \dots & A_{n 1} v_{1}^{n} + \dots + A_{n n} v_{n}^{n} \end{array}) \\ = & det ((\begin{array}{c} A_{11} & \dots & A_{1 n} \\ ⋮ & ⋱ & ⋮ \\ A_{n 1} & \dots & A_{n n} \end{array}) (\begin{array}{c} v_{1}^{1} & \dots & v_{1}^{n} \\ ⋮ & ⋱ & ⋮ \\ v_{n}^{1} & \dots & v_{n}^{n} \end{array})) \\ = & (det A) det (\begin{array}{c} v_{1}^{1} & \dots & v_{1}^{n} \\ ⋮ & ⋱ & ⋮ \\ v_{n}^{1} & \dots & v_{n}^{n} \end{array}) \\ = & (det A) d x_{1} \land \dots \land d x_{n} (v^{1}, \dots, v^{n}), \end{aligned}

which concludes the proof.

🔗

This is all very nice, but so far we only looked at basic

k

-forms, and linear maps

ϕ : R^{m} \to R^{n} .

How do we generalize this to general differential

k

-forms on

U \subseteq R^{n},

and to smooth functions

ϕ : V \to U

with

V \subseteq R^{m} ?

The idea is simple. For any point

t \in V,

the Jacobian matrix of

ϕ

(i.e. the matrix of first partial derivatives), also called the “total derivative of

ϕ

”, provides a linear map

R^{m} \to R^{n} .

This is called the “differential” or “total derivative” of the smooth function

ϕ :

in fancier differential geometry, one would say that it is a linear map from the tangent space of

V

t

to the tangent space of

U

ϕ (t) .

🔗

In other words, given a smooth function

ϕ : V \to U,

if we write

ϕ (t) = (x_{1} (t), \dots, x_{n} (t)),

we can construct the pullback of

k

-forms exactly as above, but with the specific linear map

R^{m} \to R^{n}

given by the Jacobian matrix:

A = (a_{i j}) = (\frac{\partial x_{i}}{\partial t_{j}}) .

🔗

Then we see that we recover all the formulae that we obtained previously, and that our three fundamental properties are satisfied! The pullback of a basic one-form becomes

ϕ^{*} (d x_{i}) = \frac{\partial x_{i}}{\partial t_{1}} d t_{1} + \dots + \frac{\partial x_{i}}{\partial t_{n}} d t_{n},

🔗

as before. Property 1 is obviously satisfied by definition. Lemma 4.7.11 becomes Property 2, and Corollary 4.7.12 becomes our general formula for the pullback of

k

-forms in Lemma 4.7.1. It is easy to check that Property 3 is satisfied. Finally, we obtain a general proof of Lemma 4.7.7 on

R^{n},

as this is simply Lemma 4.7.13. Neat!

🔗

Exercises 4.7.4 Exercises

🔗

1.

Consider the basic two-form $d x \land d y$ on $R^{2},$ and the polar coordinate transformation $α : R^{2} \to R^{2}$ with

α (r, θ) = (r \cos θ, r \sin θ) .

Show by explicit calculation that

α^{*} (d x \land d y) = r d r \land d θ = (det J_{α}) d r \land d θ .

Solution.

Let us start by calculating the pullback two-form. We get:

\begin{aligned} α^{*} (d x \land d y) = & (\cos θ d r - r \sin θ d θ) \land (\sin θ d r + r \cos θ d θ) \\ = & r \cos^{2} θ d r \land d θ - r \sin^{2} θ d θ \land d r \\ = & r (\cos^{2} θ + \sin^{2} θ) d r \land d θ \\ = & r d r \land d θ . \end{aligned}

Next, we show that $det J_{α} = r .$ By definition of the Jacobian matrix, we have:

\begin{aligned} det J_{α} = & det (\begin{array}{c} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial θ} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial θ} \end{array}) \\ = & det (\begin{array}{c} \cos θ & - r \sin θ \\ \sin θ & r \cos θ \end{array}) \\ = & r \cos^{2} θ + r \sin^{2} θ \\ = & r . \end{aligned}

Therefore,

α^{*} (d x \land d y) = r d r \land d θ = (det J_{α}) d r \land d θ

as claimed.

🔗

2.

Let

ω = (x^{2} + y^{2} + z^{2}) d x \land d y \land d z

on $R^{3},$ and $α : R^{3} \to R^{3}$ the spherical transformation

α (r, θ, ϕ) = (r \sin (θ) \cos (ϕ), r \sin (θ) \sin (ϕ), r \cos (θ)) .

Show by explicit calculation that
$α^{*} (d x \land d y \land d z) = r^{2} \sin (θ) d r \land d θ \land d ϕ = (det J_{α}) d r \land d θ \land d ϕ .$
Use this to find $α^{*} ω .$

Solution.

We calculate the pullback:

$\begin{aligned} α^{*} (d x \land d y \land d z) = & (\sin (θ) \cos (ϕ) d r + r \cos (θ) \cos (ϕ) d θ - r \sin (θ) \sin (ϕ) d ϕ) \\ \land (\sin (θ) \sin (ϕ) d r + r \cos (θ) \sin (ϕ) d θ + r \sin (θ) \cos (ϕ) d ϕ) \\ \land (\cos (θ) d r - r \sin (θ) d θ) \\ = & (r^{2} \sin^{3} (θ) \cos^{2} (ϕ) + r^{2} \sin (θ) \cos^{2} (θ) \cos^{2} (ϕ) + r^{2} \sin^{3} (θ) \sin^{2} (ϕ) \\ + r^{2} \sin (θ) \cos^{2} (θ) \sin^{2} (ϕ)) d r \land d θ \land d ϕ \\ = & r^{2} (\sin^{3} (θ) + \sin (θ) \cos^{2} (θ)) d r \land d θ \land d ϕ \\ = & r^{2} \sin (θ) d r \land d θ \land d ϕ . \end{aligned}$

Next we show that this $det J_{α} = r^{2} \sin (θ) .$ By definition of the Jacobian matrix, we get:

$\begin{aligned} det J_{α} = & det (\begin{array}{c} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial θ} & \frac{\partial x}{\partial ϕ} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial θ} & \frac{\partial y}{\partial ϕ} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial θ} & \frac{\partial z}{\partial ϕ} \end{array}) \\ = & det (\begin{array}{c} \sin (θ) \cos (ϕ) & r \cos (θ) \cos (ϕ) & - r \sin (θ) \sin (ϕ) \\ \sin (θ) \sin (ϕ) & r \cos (θ) \sin (ϕ) & r \sin (θ) \cos (ϕ) \\ \cos (θ) & - r \sin (θ) & 0 \end{array}) \\ = & r^{2} \sin (θ) \cos^{2} (θ) \cos^{2} (ϕ) + r^{2} \sin^{3} (θ) \sin^{2} (ϕ) + r^{2} \sin^{3} (θ) \cos^{2} (ϕ) \\ + r^{2} \sin (θ) \cos^{2} (θ) \sin^{2} (ϕ) \\ = & r^{2} \sin (θ) . \end{aligned}$

Therefore

$α^{*} (d x \land d y \land d z) = r^{2} \sin (θ) d r \land d θ \land d ϕ = (det J_{α}) d r \land d θ \land d ϕ$

as claimed.
To find $α^{*} ω$ we can use the result in (a). We get:

$\begin{aligned} α^{*} ω = & (r^{2} \sin^{2} (θ) \cos^{2} (ϕ) + r^{2} \sin^{2} (θ) \sin^{2} (ϕ) + r^{2} \cos^{2} (θ)) α^{*} (d x \land d y \land d z) \\ = & (r^{2} \sin^{2} (θ) + r^{2} \cos^{2} (θ)) r^{2} \sin (θ) d r \land d θ \land d ϕ \\ = & r^{4} \sin (θ) d r \land d θ \land d ϕ . \end{aligned}$

🔗

3.

Let

ω = (x^{2} + y^{2}) (x d y \land d z + y d z \land d x + z d x \land d y)

on $R^{3},$ and $α : R^{3} \to R^{3}$ be the cylindrical transformation

α (r, θ, w) = (r \cos θ, r \sin θ, w) .

Find $α^{*} ω .$
Show by explicit calculation that $d (α^{*} ω) = α^{*} (d ω) .$

Solution.

To simplify the calculation of the pullback, let us first calculate the pullback of the basic two-forms:

$\begin{aligned} α^{*} (d y \land d z) = & (\sin (θ) d r + r \cos (θ) d θ) \land d w \\ = & \sin (θ) d r \land d w + r \cos (θ) d θ \land d w, \end{aligned}$

$\begin{aligned} α^{*} (d z \land d x) = & d w \land (\cos (θ) d r - r \sin (θ) d θ) \\ = & - \cos (θ) d r \land d w + r \sin (θ) d θ \land d w, \end{aligned}$

and

$\begin{aligned} α^{*} (d x \land d y) = & (\cos (θ) d r - r \sin (θ) d θ) \land (\sin (θ) d r + r \cos (θ) d θ) \\ = & r \cos^{2} (θ) d r \land d θ + r \sin^{2} (θ) d r \land d θ \\ = & r d r \land d θ . \end{aligned}$

We also observe that

$α^{*} (x^{2} + y^{2}) = r^{2} \cos^{2} (θ) + r^{2} \sin^{2} (θ) = r^{2} .$

Putting this together, we get:

$\begin{aligned} α^{*} ω = & α^{*} (x^{2} + y^{2}) α^{*} (x d y \land d z + y d z \land d x + z d x \land d y) \\ = & r^{2} (r \cos (θ) (\sin (θ) d r \land d w + r \cos (θ) d θ \land d w) \\ + r \sin (θ) (- \cos (θ) d r \land d w + r \sin (θ) d θ \land d w) + r w d r \land d θ) \\ = & r^{3} (r d θ \land d w + w d r \land d θ) . \end{aligned}$
On the one hand, we calculated in (a) the pullback $α^{*} ω = r^{3} (r d θ \land d w + w d r \land d θ) .$ We can calculate its exterior derivative:

$\begin{aligned} d (α^{*} ω) = & d (r^{4}) \land d θ \land d w + d (r^{3} w) \land d r \land d θ \\ = & (4 r^{3} + r^{3}) d r \land d θ \land d w . \\ = & 5 r^{3} d r \land d θ \land d w . \end{aligned}$

On the other hand, we can calculate first the exterior derivative of $ω .$ We get:

$\begin{aligned} d ω = & d ((x^{2} + y^{2}) x) \land d y \land d z + d ((x^{2} + y^{2}) y) \land d z \land d x + d ((x^{2} + y^{2}) z) \land d x \land d y \\ = & ((3 x^{2} + y^{2}) + (3 y^{2} + x^{2}) + (x^{2} + y^{2})) d x \land d y \land d z \\ = & 5 (x^{2} + y^{2}) d x \land d y \land d z . \end{aligned}$

We then calculate its pullback:

$\begin{aligned} α^{*} (d ω) = & 5 α^{*} (x^{2} + y^{2}) α^{*} (d x \land d y \land d z) \\ = & 5 r^{2} (\cos (θ) d r - r \sin (θ) d θ) \land (\sin (θ) d r + r \cos (θ) d θ) \land d w \\ = & 5 r^{2} (r \cos^{2} (θ) + r \sin^{2} (θ)) d r \land d θ \land d w \\ = & 5 r^{3} d r \land d θ \land d w . \end{aligned}$

We conclude that

$d (α^{*} ω) = α^{*} (d ω),$

as claimed.

🔗

4.

Let

ω = z e^{x y} d x \land d y

on $R^{3},$ and $ϕ : (R_{\neq 0})^{2} \to R^{3}$ with:

ϕ (u, v) = (\frac{u}{v}, \frac{v}{u}, u v) .

Find $ϕ^{*} ω .$

Solution.

We calculate the pullback:

\begin{aligned} ϕ^{*} ω = & u v e^{\frac{u}{v} \frac{v}{u}} (\frac{1}{v} d u - \frac{u}{v^{2}} d v) \land (- \frac{v}{u^{2}} d u + \frac{1}{u} d v) \\ = & u v e (\frac{1}{u v} - \frac{1}{u v}) d u \land d v \\ = & 0. \end{aligned}

🔗

5.

Show that the pullback of an exact form is always exact.

Solution.

Let $ω$ be an exact $k$ -form on $U \subseteq R^{n},$ and let $ϕ : V \to U$ be a smooth function for $V \subseteq R^{m} .$ We want to prove that the pullback of $ω$ is exact.

Since $ω$ is exact, we know that there exists a $(k - 1)$ -form $η$ on $U$ such that $ω = d η .$ Then

ϕ^{*} ω = ϕ^{*} (d η) = d (ϕ^{*} η),

where we used the fact that the pullback commutes with the exterior derivative (see Lemma 4.7.4). Therefore, the $k$ -form $ϕ^{*} ω$ on $V$ is the exterior derivative of a $(k - 1)$ -form $ϕ^{*} η$ on $V,$ and hence it is exact.

🔗

6.

Let $ω$ be a $k$ -form on $R^{n},$ and $I d : R^{n} \to R^{n}$ be the identity function defined by $I d (x_{1}, \dots, x_{n}) = (x_{1}, \dots, x_{n}) .$ Show that

I d^{*} ω = ω .

Solution.

We can write a general $k$ -form on $R^{n}$ as

ω = \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} f_{i_{1} \dots i_{k}} d x_{i_{1}} \land \dots \land d x_{i_{k}}

for some smooth functions $f_{i_{1} \dots i_{k}} : R^{n} \to R .$ We know that $I d^{*} (d x_{i}) = d x_{i}$ for all $i \in {1, \dots, n},$ by definition of the identity map. Similarly, for any function $f : R^{n} \to R,$ $I d^{*} f = f .$ As a result, we get:

\begin{aligned} I d^{*} ω = & \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} I d^{*} (f_{i_{1} \dots i_{k}}) I d^{*} (d x_{i_{1}}) \land \dots \land I d^{*} (d x_{i_{k}}) \\ = & \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} f_{i_{1} \dots i_{k}} d x_{i_{1}} \land \dots \land d x_{i_{k}} \\ = & ω . \end{aligned}

🔗

7.

Let $ω$ be a $k$ -form on $U \subseteq R^{n} .$ Let $ϕ : V \to U$ and $α : W \to V$ be smooth functions, with $V \subseteq R^{m}$ and $W \subseteq R^{ℓ} .$ Show that

(ϕ \circ α)^{*} ω = α^{*} (ϕ^{*} ω) .

In other words, it doesn't matter whether we pullback in one or two steps in the chain of maps

W \overset{α}{\to} V \overset{ϕ}{\to} U .

Solution.

We can write a general $k$ -form on $U$ as

ω = \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} f_{i_{1} \dots i_{k}} d x_{i_{1}} \land \dots \land d x_{i_{k}}

for some smooth functions $f_{i_{1} \dots i_{k}} : U \to R .$ On the one hand, the pullback by $ϕ \circ α$ is

(ϕ \circ α)^{*} ω = \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} (ϕ \circ α)^{*} (f_{i_{1} \dots i_{k}}) (ϕ \circ α)^{*} (d x_{i_{1}}) \land \dots \land (ϕ \circ α)^{*} (d x_{i_{k}}) .

On the other hand, pulling back in two steps, we get:

α^{*} (ϕ^{*} ω) = \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} α^{*} (ϕ^{*} f_{i_{1} \dots i_{k}}) α^{*} (ϕ^{*} d x_{i_{1}}) \land \dots \land α^{*} (ϕ^{*} d x_{i_{k}}) .

So all we have to show is that

(ϕ \circ α)^{*} f = α^{*} (ϕ^{*} f)

for any smooth function $f : U \to R,$ and

(ϕ \circ α)^{*} d x_{i} = α^{*} (ϕ^{*} d x_{i})

for any $i \in {1, \dots, n} .$

First, for any function $f : U \to R,$

(ϕ \circ α)^{*} f = f \circ ϕ \circ α,

while

α^{*} (ϕ^{*} f) = α^{*} (f \circ ϕ) = f \circ ϕ \circ α .

Thus

(ϕ \circ α)^{*} f = α^{*} (ϕ^{*} f) .

As for the basic one-forms, let us introduce further notation for the maps $ϕ : V \to U$ and $α : W \to V .$ Let us write $z = (z_{1}, \dots, z_{ℓ})$ for coordinates on $W;$ $y = (y_{1}, \dots, y_{m})$ for coordinates on $V;$ and $x = (x_{1}, \dots, x_{n})$ for coordinates on $U .$ We write $ϕ (y) = (ϕ_{1} (y, \dots, ϕ_{n} (y)),$ and $α (z) = (α_{1} (z), \dots, α_{m} (z)) .$ Then, we have:

ϕ^{*} d x_{i} = \sum_{a = 1}^{m} \frac{\partial ϕ_{i}}{\partial y_{a}} d y_{a},

and

α^{*} (ϕ^{*} d x_{i}) = \sum_{b = 1}^{ℓ} \sum_{a = 1}^{m} \frac{\partial ϕ_{i}}{\partial y_{a}} |_{y = α (z)} \frac{\partial α_{a}}{\partial z_{b}} d z_{b} .

On the other hand, if we pullback by the composition of the maps, we get:

(ϕ \circ α)^{*} d x_{i} = \sum_{b = 1}^{ℓ} \frac{\partial (ϕ \circ α)_{i}}{\partial z_{b}} d z_{b} .

But the equality

\frac{\partial (ϕ \circ α)_{i}}{\partial z_{b}} = \sum_{a = 1}^{m} \frac{\partial ϕ_{i}}{\partial y_{a}} |_{y = α (z)} \frac{\partial α_{a}}{\partial z_{b}}

is precisely the chain rule for multi-variable functions (written in terms of composition of functions). Thus we conclude that

α^{*} (ϕ^{*} d x_{i}) = (ϕ \circ α)^{*} d x_{i} .

Putting all of this together, we conclude that

(ϕ \circ α)^{*} ω = α^{*} (ϕ^{*} ω),

which is the statement of the question.

Section 4.7 The pullback of a k-form

Objectives

Subsection 4.7.1 The pullback of a k-form: an axiomatic approach

Lemma 4.7.1. The pullback of a k-form.

Proof.

Example 4.7.2. The pullback of a two-form.

Example 4.7.3. The pullback of a three-form.

Lemma 4.7.4. The pullback commutes with the exterior derivative.

Proof.

Subsection 4.7.2 The pullback of a top form in Rn and the Jacobian determinant

Definition 4.7.5. The Jacobian.

Definition 4.7.6. Top form.

Lemma 4.7.7. The pullback of a top form in Rn in terms of the Jacobian determinant.

Proof for R2 and R3.

Subsection 4.7.3 The pullback of a k-form: an algebraic approach (optional)

Definition 4.7.8. The pullback of a basic one-form with respect to a linear map.

Lemma 4.7.9. An explicit formula for the pullback of a basic one-form.

Proof.

Definition 4.7.10. The pullback of a basic k-form with respect to a linear map.

Lemma 4.7.11. The pullback commutes with the wedge product.

Proof.

Corollary 4.7.12. An explicit formula for the pullback of a basic k-form.

Lemma 4.7.13. The pullback of a basic n-form in Rn.

Proof.

Exercises 4.7.4 Exercises

1.

2.

3.

4.

5.

6.

7.

Section 4.7 The pullback of a $k$ -form

Subsection 4.7.1 The pullback of a $k$ -form: an axiomatic approach

Lemma 4.7.1. The pullback of a $k$ -form.

Subsection 4.7.2 The pullback of a top form in $R^{n}$ and the Jacobian determinant

Lemma 4.7.7. The pullback of a top form in $R^{n}$ in terms of the Jacobian determinant.

Proof for $R^{2}$ and $R^{3}$ .

Subsection 4.7.3 The pullback of a $k$ -form: an algebraic approach (optional)

Definition 4.7.10. The pullback of a basic $k$ -form with respect to a linear map.

Corollary 4.7.12. An explicit formula for the pullback of a basic $k$ -form.

Lemma 4.7.13. The pullback of a basic $n$ -form in $R^{n}$ .