Pull back of $0$-tensor field

differential-geometrymanifoldssmooth-manifolds

Let $F:M\to N$ be a smooth map between smooth manifolds. Let $f$ be real value continuous function on $N$.

Prove that the pull back map has the relationship: $F^*f = f\circ F$

Where the pull back map between covariant k-tensor field for each point $p$ on manifold is defined as $(F^*A)_p(v_1,…,v_k) = A_{F(k)}(dF_p(v_1),…,dF_p(v_k)).$

Since real continuous function is $0$-tensor field then $k = 0$ is empty so I don't know how to deal with this case?

Best Answer

Ignoring the input vectors $v_1,...,v_k$, your definition of the pull-back becomes:

$$(F^* A)_p = A_{F(p)}.$$

If you identify 0-forms with functions (i.e. $\omega_p = \omega(p)$), you get

$$ (F^* A)(p) = A(F(p)) \; \forall p \in M \implies F^*A = A \circ F. $$

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[Math] Pulling back vector fields

To pull back vector fields on $N$ it suffices for $F : M \to N$ to be a local diffeomorphism (at each point of $M$):

If $Y$ is a smooth vector field on $\operatorname{Im} F$ (i.e. the image of $F$, which is an open subset of $N$ since $F : M \to N$ is an open map) then you can define a vector field $X$ on $M$ by $m \in M \mapsto \left( \operatorname{T}_m F \right)^{-1}(Y_{F(m)})$, which is well-defined since each $\operatorname{T}_m F : \operatorname{T}_m M \to \operatorname{T}_{F(m)} N$ is a bijection. $X$ will be smooth since for any $m \in M$ we can select an open (in $M$) neighbourhood $U$ of $m$ such that $F\big\vert_U : U \to F(U)$ is a diffeomorphism onto the open subset $F(U)$ of $N$. Note that $X\big\vert_U$ is equal to the pushforward $X\big\vert_U = \left( F\big\vert_U^{-1} \right)_{*}\left( Y\big\vert_{F(U)} \right) \circ F\big\vert_U$, which is smooth since it is a composition of the smooth maps $$U \xrightarrow{F\vert_U} F(U) \xrightarrow{Y\vert_{F(U)}} \operatorname{T}_{F(U)} N \xrightarrow{\left( F\vert_U^{-1} \, \right)_{*}} \operatorname{T}_{U} M$$ so that $X$ is smooth around each point of $M$.

Edit: Using the above result, by going into coordinates it is then straightforward to show that if $F : M \to N$ is an immersion and $Y_{F(m)} \in \operatorname{Im} \left( \operatorname{T}_m F \right)$ for all $m \in M$ then the map $X : M \to \operatorname{T} M$ defined by $m \mapsto \left( \operatorname{T}_m F \right)^{-1}(Y_{F(m)})$ is a well-defined smooth vector field on $M$ that can be pushed-forward to $Y\big\vert_{\operatorname{Im} F}$. Note that if $\operatorname{Im} F$ is dense in $N$ then the continuous map $Y$ is completely determined by $Y\big\vert_{\operatorname{Im} F}$ so that in this sense, $X$ could be viewed as "essentially" being the pullback of the entire vector field $Y$.

[Math] f-related vector field

Sorry for the delay, I have been away for a few days. This calculation is horrible and I don't guarantee that this is the best way to do this. I just did it in coordinates and it seems to work.

We want to show that

$$f_*[X,Y]=[f_*(X),f_*(Y)]$$

We will show this by showing that both sides act the same way on functions on $N$ and so must be the same vector field.

Let $h:N\to\mathbb R$, we have by definition

$$f_*([X,Y])(h)=[X,Y](h\circ f)$$

We use coordinates $x^i$ for $M$ and $y^m$ for $N$. We have \begin{equation} \begin{aligned}f_*([X,Y])(h)&=[X,Y](h\circ f)(x)\\ &=X^i\frac{\partial}{\partial x^i}\left(Y^j\frac{\partial}{\partial x^j}(h\circ f)\right)(x)-Y^i\frac{\partial}{\partial x^i}\left(X^j\frac{\partial}{\partial x^j}(h\circ f)\right)(x)\\ &=X^i\frac{\partial}{\partial x^i}\left(Y^j\frac{\partial h}{\partial y^m}(f(x))\frac{\partial y^m}{\partial x^i}(x)\right)-Y^i\frac{\partial}{\partial x^i}\left(X^j\frac{\partial h}{\partial y^m}(f(x))\frac{\partial y^m}{\partial x^i}(x)\right)\\ &=X^i\frac{\partial Y^j}{\partial x^i}\frac{\partial h}{\partial y^m}(f(x))\frac{\partial y^m}{\partial x^i}(x)-Y^i\frac{\partial X^j}{\partial x^i}\frac{\partial h}{\partial y^m}(f(x))\frac{\partial y^m}{\partial x^i}(x) \end{aligned} \end{equation}

While going from the other end \begin{equation} \begin{aligned} \phantom a[f_*(X),f_*(Y)](h)&=f_*(X)(f_*(Y)(h))-f_*(X)(f_*(Y)(h))\\ &=f_*(X)\left(Y^j\frac{\partial y^m}{\partial x^j}(x)\frac{\partial h}{\partial y^m}(f(x))\right)-f_*(Y)\left(X^j\frac{\partial y^m}{\partial x^j}(x)\frac{\partial h}{\partial y^m}(f(x))\right)\\ &=X^iY^j\frac{\partial y^m}{\partial x^j}(x)\frac{\partial y^n}{\partial x^i}(x)\frac{\partial^2 h}{\partial y^m\partial y^n}(f(x))+X^i\frac{\partial Y^j}{\partial x^m}\frac{\partial y^m}{\partial x^j}(x)\frac{\partial x^m}{\partial y^n}(f(x))\frac{\partial y^n}{\partial x^i}(x)\frac{\partial h}{\partial y^m}(f(x))\\ &\phantom=-Y^iX^j\frac{\partial y^m}{\partial x^j}(x)\frac{\partial y^n}{\partial x^i}(x)\frac{\partial^2 h}{\partial y^m\partial y^n}(f(x))-Y^i\frac{\partial X^j}{\partial x^m}\frac{\partial y^m}{\partial x^j}(x)\frac{\partial x^m}{\partial y^n}(f(x))\frac{\partial y^n}{\partial x^i}(x)\frac{\partial h}{\partial y^m}(f(x))\\ &=X^i\frac{\partial Y^j}{\partial x^i}\frac{\partial y^m}{\partial x^j}(x)\frac{\partial h}{\partial y^m}(f(x))-Y^i\frac{\partial X^j}{\partial x^i}\frac{\partial y^m}{\partial x^j}(x)\frac{\partial h}{\partial y^m}(f(x))\\ \end{aligned} \end{equation}

We have now shown that both sides are the same. In this derivation I have used the chain rule as well as the Jacobian, some cancellations and the fact that $\frac{\partial y^m}{\partial x^j}(f(x))\frac{\partial x^k}{\partial y^m}(x)=\delta^k_j$.

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[Math] Pulling back vector fields

[Math] f-related vector field

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