Pullback to a measure zero submanifold

Question

Suppose that $M$ is a manifold and $N$ is a measure zero submanifold of $M$. Are we allowed to define pullback/pushforward to/from this submanifold? If we are, then I'm having difficulty understanding the pullback of let's say metric tensor to a point! I'm asking this since I saw in a paper (by two physicists) that the authors defined pullback of a two-form from the space of all functions satisfying a set of specific boundary conditions, $\mathcal{C}$, to the space of solutions of a differential equation with those boundary conditions, $\mathcal{P}$ (which I think is a measure zero subspace of $\mathcal{C}$).

Answer 1

A covariant tensor can be pulled back along any smooth map (restriction to a submanifold being the special case of pulling back along the inclusion map). Indeed, suppose $\omega$ is a $k$-ary covariant tensor on $M$ and $f:N\to M$ is smooth. This means that for each $p\in M$, $\omega_p$ is a multilinear map $(T_pM)^k\to\mathbb{R}$, which varies smoothly with $p$. We can then define the pullback $f^*\omega$ by $$(f^*\omega)_q(v_1,\dots,v_k)=\omega_{f(q)}(df_q(v_1),\dots,df_q(v_n)$$ for $q\in N$ and $v_1,\dots,v_k\in T_qN$. In other words, to evaluate $f^*\omega$ on some tangent vectors, we push those tangent vectors forwards by $f$ and then evaluate $\omega$ on them.

In particular, then, we can pull back a $k$-form on $M$ to get a $k$-form on $N$. Or, we can pull back a metric on $M$ to get a symmetric bilinear form on each tangent space of $N$, though they will be degenerate unless $f$ is an immersion (since they will vanish on any tangent vectors on $N$ whose pushforwards to $M$ are $0$).

Very explicitly, in the case that $N$ is a submanifold of $M$ (and $f$ is the inclusion map), we pull a $k$-form on $M$ back to $N$ by just restricting it to the tangent space of $N$ at each point. Similarly, given a metric on $M$, we pull it back to $N$ by just restricting the metric to the tangent spaces of $N$.

Note that if $N$ is a point, then its tangent space is trivial and so any tensor on it will be trivial, including any metric or differential form you get by pulling back along some map.