Differential Geometry – Visualizing Commutator of Two Vector Fields

differential-geometrymultivariable-calculus

I'm reading a book on calculus, the part about vector fields on
manifolds. It's a nice book, but with a severe drawback — it has no pictures.

I like how vectors are treated algebraically, as derivatives over a local ring (ring of germs). But I still want to use "geometrical" view on vector fields.

The problem is I can't imagine vector field "multiplication" as a composition of derivatives. And thus I can't picture commutator of two
vector fields.

Has anybody here got pictures too help me?

Best Answer

In Gauge fields, knots and gravity by J. Baez and J. P. Muniain authors got the following two pictures to visualize commutator:
enter image description here

I hope it helps you.

Related Solutions

How to Derive Formulas for Divergence and Curl

The complete answer is given on pages 22-27 of my 2011 vector calculus notes. I think many good calculus text include these heuristic arguments, I found them in Thomas' calculus a few editions back. Long story short, what you should really do to understand is to prove Greene's and Stokes' Theorems, this will give you deeper insight into the nature of your question. Let me summarize the method here:

The flux of $\langle P,Q \rangle$ through a little rectangle with corners $(x,y), (x+\triangle x,y), (x,y+\triangle y), (x+\triangle x,y+\triangle y)$ is easily calculated by multiplying components of $\langle P,Q \rangle$ with width ($\triangle x$) and height ($\triangle y$) with the values of the components at the corners. Then divide by the area $\triangle x \triangle y$ of the rectangle and pass to the limits $\triangle x,\triangle y \rightarrow 0$ to obtains partial derivatives which we recognize as the divergence of $\langle P,Q \rangle$. It follows that, $\nabla \cdot \langle P,Q \rangle$ measures the flux area density of the vector field $\langle P,Q \rangle$. The extension to three variables is including in my notes where I discuss a sketch of the proof of Gauss' Theorem.

rectangle to motivate divergence definition

The curl is seen from doing a similar argument to calculate the circulation of $\langle P,Q \rangle$ around a little rectangle with corners $(x,y), (x+\triangle x,y), (x,y+\triangle y), (x+\triangle x,y+\triangle y)$. To find the circulation from $(x,y)$ to $(x+\triangle x,y)$ we multiply $P(x,y) \triangle x$ as the path is horizontal so only the $x$-component lines up with the segment to contribute some circulation. If you take into account the directions of each segment and divide by the area $\triangle x \triangle y$ of the rectangle and then pass to the limit $\triangle x,\triangle y \rightarrow 0$ and the $z$-component of the curl appears. It follows that, $\nabla \times \langle P,Q,0 \rangle$ measures the circulation area density of the vector field $\langle P,Q,R \rangle$ in the $z$-direction. To extend to a three component vector field we need a vector to describe the possible circulations along a space curve in all three directions.

rectangle to motivate the curl

All of this said, I really would rather give the less helpful answer that $d$ is the natural exterior derivative operation on the exterior algebra of $\mathbb{R}^3$ and it is just the case that: 1. $df = \omega_{\nabla f}$, 2. $d\omega_{\vec{F}} = \Phi_{\nabla \times \vec{F}}$ and 3. $d \Phi_{\vec{G}} = \nabla \cdot G dx \wedge dy \wedge dz$

where $\omega_{\langle a,b,c \rangle} = adx+bdy+cdz$ and $\Phi_{\langle a,b,c \rangle} = ady \wedge dz+bdz \wedge dz+cdx \wedge dy$. Therefore, gradient, curl and divergence are just different levels of the cohomological operator which ultimately reveals the deeper shape of space.

Differential Geometry – Lie Derivative Along the Commutator of Two Vector Fields

One way to prove this result is to first show it for functions on the manifold, and use this to further prove it for vector fields, co-vector fields and finally general tensor fields. The advantage of this approach as it requires the minimum of formulas.

We will need the following assumptions:

(a) The Lie derivative follows the Leibnitz rule when acting on a product of objects.

(b) $ \mathcal{L}_{X} (f) = X(f) $

Part 1 - Show for a function, f: Using (b) we have $$ \mathcal{L}_X \mathcal{L}_Y f = \mathcal{L}_X ( Y(f)) = X(Y(f)) $$

In a coordinate basis this is:

$$ X^{\rho} \partial_{\rho} (Y^{\sigma} \partial_{\sigma} f) = X^{\rho} ( \partial_{\rho} Y^{\sigma}) (\partial_{\sigma} f) + X^{\rho} Y^{\sigma} (\partial_{\rho} \partial_{\sigma} f) $$

Because $ \partial_{\rho} \partial_{\sigma} f $ is symmetric in $\rho \leftrightarrow \sigma $ then the second term on the RHS is symmetric in $ X \leftrightarrow Y $. Therefore:

$$ \mathcal{L}_X \mathcal{L}_Y f - \mathcal{L}_Y \mathcal{L}_X f = X^{\rho} ( \partial_{\rho} Y^{\sigma}) (\partial_{\sigma} f) - Y^{\rho} ( \partial_{\rho} X^{\sigma}) (\partial_{\sigma} f) = (X^{\rho} ( \partial_{\rho} Y^{\sigma}) - X^{\rho} ( \partial_{\rho} Y^{\sigma}))(\partial_{\sigma} f) $$

$$ = [X,Y]^{\sigma} \partial_{\sigma} f = [X,Y](f) = \mathcal{L}_{[X,Y]} f $$

Part 2: Show for a vector field, $ T = V \in TM $

Using the result from Part 1 we have:

$$ \mathcal{L}_{[X,Y]} (V(f)) = \mathcal{L}_{X} \mathcal{L}_{Y} (V(f)) - \mathcal{L}_{Y} \mathcal{L}_{X} (V(f)) $$

Using the Leibnitz rule:

$$ \mathcal{L}_{X} \mathcal{L}_{Y} (V(f)) = \mathcal{L}_{X} (( \mathcal{L}_{Y} V)f) + \mathcal{L}_{X}( V(\mathcal{L}_{Y} f) ) = (\mathcal{L}_{X} \mathcal{L}_{Y} V)(f) + V(\mathcal{L}_{X} \mathcal{L}_{Y}f) + (\mathcal{L}_{X} V)(\mathcal{L}_{Y} f) + (\mathcal{L}_{Y} V)(\mathcal{L}_{X} f) $$

The last 2 terms taken together are symmetric in $ X \leftrightarrow Y $. We therefore have that

$$ \mathcal{L}_{[X,Y]} (V(f)) = \mathcal{L}_{X} \mathcal{L}_{Y} (V(f)) - \mathcal{L}_{Y} \mathcal{L}_{X} (V(f)) = (\mathcal{L}_{X} \mathcal{L}_{Y} V - \mathcal{L}_{Y} \mathcal{L}_{X} V)(f) + V(\mathcal{L}_{X} \mathcal{L}_{Y} f - \mathcal{L}_{Y} \mathcal{L}_{X}f) $$

But we can also apply the Leibnitz rule to $\mathcal{L}_{[X,Y]} (V(f))$ to give:

$$ \mathcal{L}_{[X,Y]} (V(f)) = (\mathcal{L}_{[X,Y]} V)(f) + V(\mathcal{L}_{[X,Y]} f)) = (\mathcal{L}_{[X,Y]} V)(f) + V(\mathcal{L}_{X} \mathcal{L}_{Y} f - \mathcal{L}_{Y} \mathcal{L}_{X}f))$$

Comparing these 2 expressions we get that:

$$ (\mathcal{L}_{[X,Y]} V)(f) = (\mathcal{L}_{X} \mathcal{L}_{Y} V - \mathcal{L}_{Y} \mathcal{L}_{X} V)(f) $$

This is true for any function hence we have shown the identity for a vector field.

Part 3 - Show for a covector field, $ T = \eta \in T^{*}M $

We consider the action of $ \mathcal{L}_{[X,Y]} $ on the function $ \eta (V) $ where $ \eta $ is a covector field and V is a vector field. This works exactly the same as in part 2 giving us:

$$(\mathcal{L}_{[X,Y]} \eta)(V) = (\mathcal{L}_{X} \mathcal{L}_{Y} \eta - \mathcal{L}_{Y} \mathcal{L}_{X} \eta)(V) $$

This is true for any vector field hence the result is shown for covector fields.

Part 4 - General Tensor field, T

We will now consider the action of $ \mathcal{L}_{[X,Y]} $ on the function $T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s)$, where T is an (r,s) rank tensor, $ X_i $ is a vector field, and $ \eta_i $ is a covector field. This step proceeds similarly to before.

By the result from Part 1:

$$ \mathcal{L}_{[X,Y]}T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) = (\mathcal{L}_{X} \mathcal{L}_{Y} - \mathcal{L}_{Y} \mathcal{L}_{X})T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) $$

Similarly to Part 2 $ \mathcal{L}_{X} \mathcal{L}_{Y} $ acting on $ T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) $ gives terms symmetric in $ X \leftrightarrow Y $ when the Lie derivatives hit seperate terms. These terms will cancel with terms contained in $ - \mathcal{L}_{Y} \mathcal{L}_{X} T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) $. Therefore we are left with:

$$ (\mathcal{L}_{X} \mathcal{L}_{Y} - \mathcal{L}_{Y} \mathcal{L}_{X})T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) = (\mathcal{L}_{X} \mathcal{L}_{Y}T - \mathcal{L}_{Y} \mathcal{L}_{X}T)(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T((\mathcal{L}_{X}\mathcal{L}_{Y} - \mathcal{L}_{Y} \mathcal{L}_{X})X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T(X_1,(\mathcal{L}_{X}\mathcal{L}_{Y} - \mathcal{L}_{Y} \mathcal{L}_{X})X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + \dots $$

Using the results from Parts 2 and 3 gives:

$$ \mathcal{L}_{[X,Y]}T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) =(\mathcal{L}_{X} \mathcal{L}_{Y}T - \mathcal{L}_{Y} \mathcal{L}_{X}T)(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T(\mathcal{L}_{[X,Y]}X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T(X_1,\mathcal{L}_{[X,Y]}X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + \dots $$

Lastly, the use of the Leibnitz rule gives

$$ \mathcal{L}_{[X,Y]}T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) = \mathcal{L}_{[X,Y]}T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T(\mathcal{L}_{[X,Y]}X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T(X_1,\mathcal{L}_{[X,Y]}X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + \dots $$

Comparing the 2 equalities and noting that they are true for any vector fields $ X_1,X_2,...,X_r $ and any covector fields $ \eta_1,\eta_2,...,\eta_s $ proves that the desired result holds for any general tensor field T.

Best Answer

Related Solutions

How to Derive Formulas for Divergence and Curl

Differential Geometry – Lie Derivative Along the Commutator of Two Vector Fields

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