# The Unapologetic Mathematician

## The Lie Bracket of Vector Fields

We know that any vector field $X\in\mathfrak{X}U$ can act as an endomorphism on the space $\mathcal{O}U$ of smooth functions on $U$. What happens if we act by one vector field followed by another? To really make things explicit, let’s say that $(U,x)$ is a coordinate patch, so we can write $\displaystyle Xf(p)=\sum\limits_{i=1}^nX^i(p)\frac{\partial f}{\partial x^i}\bigg\vert_p$

where $X^i$ is a coefficient function and $\frac{\partial f}{\partial x^i}\vert_p$ measures how fast $f$ is changing as we increase the $i$th coordinate function through $p$. Now if we hit this with another vector field $Y$ we find $\displaystyle YXf=\sum\limits_{i=1}^nY\left(X^i\frac{\partial f}{\partial x^i}\right)$

At each point $p$ the field $Y$ gives a vector $Y_p$, which acts as a derivation on the ring of smooth functions at $p$. That is \displaystyle\begin{aligned}YXf&=\sum\limits_{i=1}^nY\left(X^i\right)\frac{\partial f}{\partial x^i}+X^iY\left(\frac{\partial f}{\partial x^i}\right)\\&=\sum\limits_{i=1}^n\sum\limits_{j=1}^nY^j\frac{\partial X^i}{\partial x^j}\frac{\partial f}{\partial x^i}+X^iY^j\frac{\partial^2f}{\partial x^j\partial x^i}\end{aligned}

Now, this is obviously an endomorphism on $\mathcal{O}U$ since it’s the composite of two endomorphisms. But it is not a vector field, since at a given point $p$ we don’t get a derivation of the ring of smooth functions at $p$. Indeed, what happens if we give it the product of two functions? \displaystyle\begin{aligned}YX(fg)&=\sum\limits_{i=1}^n\sum\limits_{j=1}^nY^j\frac{\partial X^i}{\partial x^j}\frac{\partial(fg)}{\partial x^i}+X^iY^j\frac{\partial}{\partial x^j}\frac{\partial(fg)}{\partial x^i}\\&=\sum\limits_{i=1}^n\sum\limits_{j=1}^nY^j\frac{\partial X^i}{\partial x^j}\left(\frac{\partial f}{\partial x^i}g+f\frac{\partial g}{\partial x^i}\right)+X^iY^j\frac{\partial}{\partial x^j}\left(\frac{\partial f}{\partial x^i}g+f\frac{\partial g}{\partial x^i}\right)\\&=\sum\limits_{i=1}^n\sum\limits_{j=1}^nY^j\frac{\partial X^i}{\partial x^j}\left(\frac{\partial f}{\partial x^i}g+f\frac{\partial g}{\partial x^i}\right)+X^iY^j\left(\frac{\partial^2f}{\partial x^j\partial x^i}g+\frac{\partial f}{\partial x^i}\frac{\partial g}{\partial x^j}+\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^i}+f\frac{\partial^2g}{\partial x^j\partial x^i}\right)\\&=YX(f)g+fYX(g)+\sum\limits_{i=1}^n\sum\limits_{j=1}^nX^iY^j\left(\frac{\partial f}{\partial x^i}\frac{\partial g}{\partial x^j}+\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^i}\right)\end{aligned}

We’ve got a bunch of terms left over at the end! But one thing is nice about it: the leftover terms are symmetric between $X$ and $Y$: \displaystyle\begin{aligned}XY(fg)&=XY(f)g+fXY(g)+\sum\limits_{i=1}^n\sum\limits_{j=1}^nY^iX^j\left(\frac{\partial f}{\partial x^i}\frac{\partial g}{\partial x^j}+\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^i}\right)\\&=XY(f)g+fXY(g)+\sum\limits_{i=1}^n\sum\limits_{j=1}^nX^iY^j\left(\frac{\partial f}{\partial x^i}\frac{\partial g}{\partial x^j}+\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^i}\right)\end{aligned}

So what would happen if instead of using the regular composition product of these endomorphisms, we used the associated Lie bracket? We’d find \displaystyle\begin{aligned}{}[X,Y](fg)=&XY(fg)-YX(fg)\\=&XY(f)g+fXY(g)+\sum\limits_{i=1}^n\sum\limits_{j=1}^nX^iY^j\left(\frac{\partial f}{\partial x^i}\frac{\partial g}{\partial x^j}+\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^i}\right)\\&-YX(f)g+fYX(g)+\sum\limits_{i=1}^n\sum\limits_{j=1}^nX^iY^j\left(\frac{\partial f}{\partial x^i}\frac{\partial g}{\partial x^j}+\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^i}\right)\\=&XY(f)g-YX(f)g+fXY(g)-fYX(g)\\=&[X,Y](f)g+f[X,Y](g)\end{aligned}

That is, the Lie bracket $[X,Y]$ of $X$ and $Y$ is another vector field! Indeed, let’s see what it looks like in coordinates: \displaystyle\begin{aligned}{}[X,Y]f=&XYf-YXf\\=&\sum\limits_{i=1}^n\sum\limits_{j=1}^nX^j\frac{\partial Y^i}{\partial x^j}\frac{\partial f}{\partial x^i}+Y^iX^j\frac{\partial^2f}{\partial x^j\partial x^i}\\&-\sum\limits_{i=1}^n\sum\limits_{j=1}^nY^j\frac{\partial X^i}{\partial x^j}\frac{\partial f}{\partial x^i}+X^iY^j\frac{\partial^2f}{\partial x^j\partial x^i}\\=&\sum\limits_{i=1}^n\left(\sum\limits_{j=1}^nX^j\frac{\partial Y^i}{\partial x^j}-Y^j\frac{\partial X^i}{\partial x^j}\right)\frac{\partial f}{\partial x^i}\end{aligned}

where we can cancel off the two second partial derivatives because we’re assuming that $f$ is “smooth”, which in this case entails “has mixed second partial derivatives which commute” in any local coordinate system.

And so we might appropriately write $\displaystyle[X,Y]^i=\sum\limits_{j=1}^nX^j\frac{\partial Y^i}{\partial x^j}-Y^j\frac{\partial X^i}{\partial x^j}$

Of course, even where we don’t have local coordinates we can still write $[X,Y]f=XYf-YXf$ or $[X,Y]=XY-YX$ and get a vector field. We may also find it useful to write down the value of this field at a point: $[X,Y]_p=X_pY-Y_pX$. Indeed we can check that this behaves like a vector at $p$: \displaystyle\begin{aligned}{}[X,Y]_p(fg)=&X_p(Y(fg))-Y_p(X(fg))\\=&X_p(Y(f)g+fY(g))-Y_p(X(f)g+fX(g))\\=&X_p(Yf)g(p)+Yf(p)X_p(g)+X_p(f)Yg(p)+f(p)X_p(Yg)\\&-Y_p(Xf)g(p)-Xf(p)Y_p(g)-Y_p(f)Xg(p)-f(p)Y_p(Xg)\\=&X_p(Yf)g(p)+Y_p(f)X_p(g)+X_p(f)Y_p(g)+f(p)X_p(Yg)\\&-Y_p(Xf)g(p)-X_p(f)Y_p(g)-Y_p(f)X_p(g)-f(p)Y_p(Xg)\\=&\left[X_pY-Y_pX\right](f)g(p)+f(p)\left[X_pY-Y_pX\right](g)\\=&[X,Y]_p(f)g(p)+f(p)[X,Y]_p(g)\end{aligned}

And so the space $\mathfrak{X}U$ of smooth vector fields on $U$ forms a Lie subalgebra of the Lie algebra of endomorphisms of the vector space $\mathcal{O}U$.

June 2, 2011 - Posted by | Differential Topology, Topology

## 5 Comments »

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