# The Unapologetic Mathematician

## Isometries

Sorry for the delay but it’s been sort of hectic with work, my non-math hobbies, and my latest trip up to DC.

Anyway, now that we’ve introduced the idea of a metric on a manifold, it’s natural to talk about mappings that preserve them. We call such maps “isometries”, since they give the same measurements on tangent vectors before and after their application.

Now, normally there’s no canonical way to translate tensor fields from one manifold to another so that we can compare them, but we’ve seen one case where we can do it: pulling back differential forms. This works because differential forms are entirely made from contravariant vector fields, so we can pull back by using the derivative to push forward vectors and then evaluate.

So let’s get explicit: say we have a metric $g_N$ on the manifold $N$, which gives us an inner product $g_{N,q}$ on each tangent space $\mathcal{T}_qN$. If we have a smooth map $f:M\to N$, we want to use it to define a metric $f^*g_N$ on $M$. That is, we want an inner product $(f^*g_N)_p$ on each tangent space $\mathcal{T}_pM$.

Now, given vectors $v$ and $w$ in $\mathcal{T}_pM$, we can use the derivative $f_*$ to push them forward to $f_*v$ and $f_*w$ in $\mathcal{T}_{f(p)}N$. We can hit these with the inner product $g_{N,f(p)}$, defining

$\displaystyle\left[(f^*g_N)_p\right]\left(v,w\right)=g_{N,f(p)}\left(f_*v,f_*w\right)$

It should be straightforward to check that this is indeed an inner product. To be thorough, we must also check that $f^*g_N$ is actually a tensor field. That is, as we move $p$ continuously around $M$ the inner product $(f^*g_N)_p$ varies smoothly.

To check this, we will use our trick: let $(U,x)$ be a coordinate patch around $p$, giving us the basic coordinate vector fields $\frac{\partial}{\partial x^i}$ in the patch. If $(V,y)$ is a coordinate patch around $f(p)$, then we know how to calculate the derivative $f_*$ applied to these vectors:

$\displaystyle f_*\frac{\partial}{\partial x^i}=\sum\limits_{k=1}^n\frac{\partial(y^k\circ f)}{\partial x^i}\frac{\partial}{\partial y^k}$

so we can stick this into the above calculation:

\displaystyle\begin{aligned}\left[(f^*g_N)_p\right]\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)&=g_{N,f(p)}\left(f_*\frac{\partial}{\partial x^i},f_*\frac{\partial}{\partial x^j}\right)\\&=g_{N,f(p)}\left(\sum\limits_{k=1}^n\frac{\partial(y^k\circ f)}{\partial x^i}\frac{\partial}{\partial y^k},\sum\limits_{l=1}^n\frac{\partial(y^l\circ f)}{\partial x^j}\frac{\partial}{\partial y^l}\right)\\&=\sum\limits_{k,l=1}^n\frac{\partial(y^k\circ f)}{\partial x^i}\frac{\partial(y^l\circ f)}{\partial x^j}g_{N,f(p)}\left(\frac{\partial}{\partial y^k},\frac{\partial}{\partial y^l}\right)\end{aligned}

Since we assumed that $g_N$ is a metric, the evaluation on the right side is a smooth function, and thus the left side is as well. So we conclude that $f^*g_N$ is a smooth tensor field, and thus a metric.

Now if $M$ comes with its own metric $g_M$, we can ask if the pull-back $f^*g_N$ is equal to $g_M$ at each point. If it is, then we call $f$ an isometry. It’s also common to say that $f$ “preserves the metric”, even though the metric gets pulled back not pushed forward.

September 27, 2011 - Posted by | Differential Geometry, Geometry

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