pullback of an Euclidian inner product on $\mathbb{R^{2n+1}}$

Question

If $M$ is a differentiable manifold of dimension n and there is a diffeomorphism $g:M \rightarrow \mathbb{R^{2n+1}}$, and we have $\langle\cdot,\cdot\rangle$ is the Euclidian inner product on $\mathbb{R^{2n+1}}$. Then the pullback of $g$ defines a metric on $M$ by $t:=g*(\langle\cdot,\cdot\rangle)$, my question is:how is $t$ defined by? I mean can you write it as a mapping and it maps what to what...

Answer 1

Let $M,N$ be two smooth manifolds, $f:M\to N$ a smooth map and let $S$ be a tensor field on $N$ of type $(0,k)$ (so at every point it eats $k$ elements of the tangent space and spits out a real number). Then, the definition of $f^*S$ is as follows: for every $p\in M$ and every $\xi_1,\dots, \xi_k\in T_pM$, we define \begin{align} (f^*S)_p[\xi_1,\dots, \xi_p]:= S_{f(p)}[f_{*,p}(\xi_1), \dots, f_{*,p}(\xi_k)] \end{align} where $f_{*,p}:T_pM\to T_{f(p)}N$ is the induced linear mapping between tangent spaces.

Recall that $S_{f(p)}:(T_{f(p)}N)^k\to \Bbb{R}$ is a multilinear map, and we're using this to define the multilinear map $(f^*S)_p: (T_pM)^k\to\Bbb{R}$. If this is confusing, then you should certainly review the linear algebra questions of how a linear map between vector spaces induces mappings between the various tensor spaces, because all we're doing in the differential geometry setting is doing the linear algebra operations pointwise at every tangent space.

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Now, suppose $M$ is a smooth manifold, $(N,h)$ is a Riemannian manifold, and $f:M\to N$ is an immersion (i.e for every $p\in M$, $f_{*,p}:T_pM\to T_{f(p)}N$ is injective). Then, we can pull-back the Riemannian metric on $N$ (which by definition is a $(0,2)$ tensor field) to get $f^*h$, which is also a Riemannian metric (since $f$ is an immersion). By using the definition above, what this says is for every $p\in M$ and every $\xi,\eta\in T_pM$, we have \begin{align} (f^*h)_p[\xi,\eta]&= h_{f(p)}[f_{*,p}(\xi),f_{*,p}(\eta)] \end{align} Or, if you want to denote the Riemannian metric as $h=\langle \cdot,\cdot \rangle$, and the pull-back as $t=f^*(\langle\cdot ,\cdot \rangle)$then \begin{align} t_p[\xi,\eta]&=\langle f_{*,p}(\xi),f_{*,p}(\eta)\rangle_{f(p)}. \end{align}

Now, if you want the expression in components, that's also simple. Let's write $t=f^*h$, fix a point $p\in M$, a chart $(U,x)$ around $p$, and a chart $(V,y)$ around $f(p)$. Then, \begin{align} t_{ij}(p)&:= t_p\left(\frac{\partial}{\partial x^i}\bigg|_{p},\frac{\partial}{\partial x^j}\bigg|_{p}\right)\\ &=h_{f(p)}\left(\frac{\partial (y^a\circ f)}{\partial x^i}\bigg|_{p}\cdot \frac{\partial }{\partial y^a}\bigg|_{f(p)}, \frac{\partial (y^b\circ f)}{\partial x^i}\bigg|_{p}\cdot \frac{\partial }{\partial y^b}\bigg|_{f(p)} \right)\\ &= \frac{\partial (y^a\circ f)}{\partial x^i}\bigg|_{p}\frac{\partial (y^b\circ f)}{\partial x^j}\bigg|_{p} h_{ab}(f(p)) \end{align} where I've used the summation convention. If you want to omit the points of evaluation, then this can be written as \begin{align} t_{ij}&= \frac{\partial (f^*y^a)}{\partial x^i}\frac{\partial (f^*y^a)}{\partial x^i} f^*(h_{ab}) \end{align} Or, the more common way to write this is to suppress the pullback symbol by saying $y=f(x)$; this yields the memorable formula (summation enforced): \begin{align} t_{ij}&=\frac{\partial y^a}{\partial x^i}\frac{\partial y^b}{\partial x^j}h_{ab}. \end{align}

Finally, if you take $h$ to be the Euclidean metric, then in terms of cartesian coordinates (i.e the identity chart), we have $h_{ab}=\delta_{ab}$, so the above formula reduces to $t_{ij}=\sum_{a}\frac{\partial y^a}{\partial x^i}\frac{\partial y^a}{\partial x^j}$.

Answer 2

Suppose that you have two spaces $X$ and $Y,$ a metric $d$ on $Y$, and a function $f : X \to Y.$ The pullback metric is the following metric on $X$:

$$(f^*d)(x^{(1)}, x^{(2)}) = d(f(x^{(1)}), f(x^{(2)})); \quad x^{(1)}, x^{(2)} \in X$$

Thus, we define a metric on $X$ by mapping points over to $Y$ and taking the distance there.

One example is given by considering different coordinate systems. Let $M = \mathbb R^n$ be the plane with ordinary Euclidean distance, then

$$d_M((x_1, y_1), (x_2, y_2), \ldots (x^n, y^n)) = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2+\ldots +(x_{n-1}-x_n)^2}.$$