How Does the Volume Form on the Unit Sphere Relate to Its Position Vector?

[spaghetti3451]

The volume form on the unit sphere $S^{n}$ in $\mathbb{R}^{n+1}$ is given by

$$i_{{\bf r}}\ dx^{1}\wedge \dots \wedge dx^{n+1}=\sum (-1)^{i-1}x^{i}dx^{1}\wedge\dots \widehat{dx^{i}} \dots \wedge dx^{n+1}.$$

Why must the volume form $dx^{1}\wedge \dots \wedge dx^{n+1}$ act on the vector ${\bf r}$ to give the volume form on the unit sphere?

Also, how do I get the form of the volume form on the right-hand side of the equation?

 

[Orodruin]

The volume form on the unit sphere $S^{n}$ in $\mathbb{R}^{n+1}$ is given by

$$i_{{\bf r}}\ dx^{1}\wedge \dots \wedge dx^{n+1}=\sum (-1)^{i-1}x^{i}dx^{1}\wedge\dots \widehat{dx^{i}} \dots \wedge dx^{n+1}.$$

Why must the volume form $dx^{1}\wedge \dots \wedge dx^{n+1}$ act on the vector ${\bf r}$ to give the volume form on the unit sphere?

What you have is the volume form in n+1 dimensions. Taking the interior product with the position vector gives you an appropriate n-form that has exactly the properties that you are looking for when you do the pullback to the n-sphere.

Also, how do I get the form of the volume form on the right-hand side of the equation?

Do you mean how you get the left-hand side on that form? Have you tried simply applying the definition of the interior product?

 

[zinq]

"Why must the volume form dx₁∧⋯∧dx_n+1 act on the vector r to give the volume form on the unit sphere?"

This works as long as by r you mean the radial vector field of unit length.

Here's one reason for this: Suppose near some point p of the unit sphere Sⁿ in Rⁿ⁺¹ you have tangent vectors

∂/∂y₁, . . ., ∂/∂y_n​

that are orthonormal at p. Then the dual 1-forms

dy₁, . . ., dy_n​

automatically have the property that their wedge product

dy₁∧ . . . ∧dy_n​

is the volume form of Sⁿ.

This work for any Riemannian manifold, not just Sⁿ. Because the unit radial vector r together with the vectors ∂/∂y₁, . . ., ∂/∂y_n form an orthonormal basis for Rⁿ⁺¹ at the point p, the wedge product of their dual 1-forms is the volume form of Rⁿ⁺¹. That is,

dx₁∧⋯∧dx_n+1 = dr∧dy₁∧ . . . ∧dy_n​

where the meaning of dr should be clear. This implies what you asked about almost immediately. It would be good to try some simple cases in low dimensions to see this is a general principle.

 

[Orodruin]

This works as long as by r you mean the radial vector field of unit length.

The position vector always has unit length on the unit sphere.

 

[zinq]

Good point!

 

[spaghetti3451]

What you have is the volume form in n+1 dimensions. Taking the interior product with the position vector gives you an appropriate n-form that has exactly the properties that you are looking for when you do the pullback to the n-sphere.

Ok, so, to get the volume form on the unit sphere $S^{n}$, you take the interior product of the volume form in $\mathbb{R}^{n+1}$ with $\bf r$.

But, how you do simplify it to the expression $\sum (-1)^{i-1}x^{i}dx^{1}\wedge\dots \widehat{dx^{i}} \dots \wedge dx^{n+1}$?

In particular, I don't understand why $\displaystyle{{\bf r} = \sum_{i=1}^{n+1} x^{i} \frac{\partial}{\partial x^{i}}}$.

My hunch is that $\displaystyle{{\bf r} = \sum_{i=1}^{n+1} \frac{\partial}{\partial x^{i}}}$ since $\bf r$ is a unit vector and must have components which are unity. But, I don't see why my hunch is wrong.

 

[Orodruin]

The vector you quote is neither radial nor has norm one. A unit vector has norm one, not all components equal to one.

 

[spaghetti3451]

Thank you! This was helpful.