What does it mean to induce a Riemannian metric on an evolving hypersurface immersed in a Riemannian manifold?

Question

Oftentimes in some journal articles, I encounter a statement like "Let $F:M^n\times[0,T]\to N^{n+1}$ be a one-parameter family of immersions in a Riemannian manifold $(N,g)$ and let $g_t$ be the induced metric $F_t^*g$," which makes me wonder why these authors consider $g_t$ while they also consider $M_t$, where $M_t$ is defined as the image $F_t(M)$ of $M$ under $F_t:=F(\cdot,t)$ for $t\in[0,T]$. This does happen, for example, when I saw it in Geometric evolution equations for hypersurfaces (Huisken & Polden, 1996), among other reference on geometric flow. And I'm very confused: which hypersurface do they really care about, $M$ or $M_t$?

Let me assume each $F_t$ is an injective immersion to start our discussion. In this case, each $M_t$ is an immersed submanifold of $M$ such that $F_t$ is diffeomorphic onto its image, according to Proposition 5.18 in Introduction to Smooth Manifolds by John M. Lee. Now this same book guarantees me that the inclusion map $\iota_t:M_t\hookrightarrow N$ induces a Riemannian metric $\iota_t^*g$ on $M_t$ through the pull-back $\iota_t^*$ (Proposition 13.9). Then I'm wondering which to consider: $(M,g_t)$ or $(M_t,\iota_t^*g)$?

Let me push the question a little further. If $F$ evolves according to $$\frac{\partial}{\partial t}F=f\nu,\tag{1}$$ where $\nu$ is a unit normal, then it is a well-known property (at least in reference on geometric flow) that the induced volume measure $d\mu$ is related to the mean curvature $H$ by $$\frac{\partial}{\partial t}d\mu=fHd\mu.\tag{2}$$ Honestly, I don't know what a volume measure is. Is that the same thing that I met in the measure theory of a real analysis course? To avoid distraction, let us interpret $d\mu$ as a Riemannian volume form defined in Proposition 15.29 of ISM. Just recently I asked how to prove (2) in Riemannian volume forms on a family of surfaces evolving by IMCF, and Professor Deane replied with great patience, but I just couldn't get it because I don't know which to consider: $g_t$ or $\iota_t^*g$. If Huisken and Polden are considering the evolution of $g_t$, then why do they define $F_t(M)$ as $M_t$ for short?

This confusion has been lingering in my mind for a long time and has seriously kept me from progressing. I really want to resolve it. Please kindly answer my question. Thank you.

Edit. Let me present another piece of evidence that supports my question. The following is an excerpt from the book Geometric Relativity written by Dan A. Lee. Judging from the context, I find it really difficult to not believe that Theorem 4.27 talks about the image $\Sigma_t$ instead of $\Sigma$.

Answer 1

And I'm very confused: which hypersurface do they really care about, $M$ or $M_t$.

In some sense, this is really the wrong question to ask. They care about a family of Riemannian manifolds: one for each value of $t$. It is up to you how you decide to implement this. Then, for extrinsic questions, you would have to specify how these Riemannian manifolds are to ‘sit inside’ another Riemannian manifold. But, I’ll elaborate my view.

Since $F$ is a 1-parameter family of immersions, it is often easier to formulate things using $M$ and the corresponding Riemannian metric $g_t:=F_t^*g$, i.e you consider the 1-parameter family of Riemannian manifolds $(M,g_t)$. So, you keep the same underlying smooth manifold, while you vary the metric tensor. Keep in mind that the ‘geometry’ (i.e lengths and angles) is all encoded in the metric.

You mention Lee’s proposition 5.18, so let me reiterate what I already mentioned in the comments. Because you’re dealing with immersions, you better be careful with the topology. The image set $M_t:=F_t[M]$ is not necessarily an embedded submanifold of $N$ (i.e the slice-chart condition could fail), and since almost everyone is accustomed to by default think of equipping a subset (here $M_t$) of a topological space (here $N$) as being equipped with the induced subspace topology, you’re going to have to do lots of mental gymnastics if you try to visualize everything using $M_t$ (of course this can help, but don’t be misled by ‘intuition’). This is why to avoid issues, it is simpler to fix the underlying smooth manifold to be $M$, and let all the dependence be encoded in the map (namely $F$). This viewpoint is also convenient is you want to do derivatives or really calculus of any sort. Also if for whatever reason you don’t require the immersions to be injective then that seems to me to be all the more reason to avoid looking at/formulating things on the image set (apart from having a vague visualization… but really our visualization corresponds to the scenario when we have a 1-parameter family of embeddings).

Having said all of this, you of course don’t need to do things this way. Suppose we have a 1-parameter family of injective immersions. The map $F_t:M\to N$ restricts to a bijection $f_t:M\to M_t$, and the domain is a smooth manifold, so by transport of structure, you can equip the target $M_t$ with a topology and smooth structure such that $F_t$ becomes a diffeomorphism. Then, the inclusion $\iota_t:M_t\to N$ becomes an injective immersion (since $\iota_t=F_t\circ f_t^{-1}$ is the composition of an injective immersion with a diffeomorphism). So, yes you can consider the Riemannian manifold $(M_t,\iota_t^*g)$. The map $f_t$ is an isometry of the Riemannian manifold $(M,g_t)$ onto $(M_t,\iota_t^*g)$, so geometrically everything is preserved (but again, I can’t stress this enough, $M_t$ has a different topology in general than the subspace topology induced from the ambient $N$, so be very careful with ‘intuition’). In other words, every geometric statement you formulate in one way, can equivalently be rephrased using another simply by composing/pulling back/pushing forward by $f_t$ (or its inverse) appropriately (this remark applies to your edit as well).

Let me now make some other remarks. An immersion is ‘locally’ an embedding (be very careful with what this does and does not mean; Lee is careful to point out the differences), so as long as you’re only dealing with local questions in topology/geometry, it doesn’t matter which you decide to take as your underlying set. You can always formulate things one way or the other (though like I said above, for calculations, it is much more convenient to fix the set, and allow the map to vary). This is also why when talking about the extrinsic curvature (second fundamental form) although one starts out generally by only considering an immersion of one Riemannian manifold into another, we often pretend it is actually an embedded submanifold (again Lee makes comments about this in his Riemannian geometry text).

Finally, if we’re in the special case where we have a 1-parameter family of embeddings, then for visualization purposes we can freely consider the image sets $M_t=F_t[M]$ and we don’t have to worry about any caveats. Stating theorems and understanding the statements is very intuitive in this setting. But again, apart from visualization, for concrete computations, it’s better to fix the set and transfer everything into the map.

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Honestly, I don't know what a volume measure is.

See the comment for your previous question about this. Again, you can pose the same question as to where the measure $\mu_t$ is defined, either $M$ or $M_t$. Well, by means of the isometry $f_t:M\to M_t$, you can go back and forth between any formulation. Though, once again the same remarks apply regarding visualization vs computation.

Answer 2

$\newcommand\II{\operatorname{II}}$The key points are:

1. The study of geometric flows is a lot easier if the setup is for time-dependent immersions and not just time-dependent embeddings. That allows you to separate the analytic aspects (studying the local properties of the flow, including existence, uniqueness, regularity of the flow and whether for each time $t$, $F_t$ is an immersion) and the global topological aspects (whether for each $t$ $F_t(M)$ is a submanifold).
2. It is always a good idea to make the domain not dependent on a parameter. In general, if you have a time-dependent domain $D_t$ and a time-dependent map $F_t: D_t \rightarrow R$, then if $s \ne t$, then $F_s$ and $F_t$ have different domains and have nothing to do with each other, unless you assume there is a diffeomorphism $\Phi_{s,t}: D_s \rightarrow D_t$. But then you need to figure out how to define $\Phi_{s,t}$. You also now have to deal with both $F$ and $\Phi_{s,t}$, making things unnecessarily complicated.

In general, you always want to parameterize maps and not sets. So the answer to your question is that you should always use $M$ as the domain and pull everything back to $M$. The computations all become quite straightforward.

Let's start at the beginning.

Let $M$ be an abstract $n$-manifold and $N$ an $(n+1)$-manifold with a Riemannian metric $h$. Given an immersion $f: M \rightarrow N$, we can pull back the metric $h$ and obtaina metric $$ g = f^*h. $$ By definition, if $v,w \in T_pM$, then $f_*v, f_w \in T_{f(p)}N$ and $$ g(v,w) = h(f_*v,f_*w). $$

Separately, the immersion $f$ is an embedding if it is injective and the map $f: M \rightarrow f(M)$ is a homeomorphism, where $f(M) \subset N$ has the subspace topology.

If we do not assume that $f$ is an embedding, then $f(M)$ is not necessarily a submanifold. The discussion below assumes only immersions. For that reason, one cannot assume $f(M)$ is a submanifold.

If $f$ is an immersion, then for each $p$, $f_*: T_pM \rightarrow T_{f(p)}N$ is an injective linear maps and $h$ defines an inner product on $T_{f(p)}N$. Therefore, there is a uniquely (up to sign) defined $\nu(p) \in T_{f(p)}N$ that is a unit normal to $f_*T_pM$. This defines a map $$ \nu: M \rightarrow T_*N, $$ where $\nu(p) \in T_{f(p)}N$. There is also a well-defined second fundamental form, which is given by a section $\II$ of $S^2T^*M$.

A time-dependent map from $M$ to $N$ is a smooth map $$ F: [0,T) \times M \rightarrow N. $$ For convenience, we write $$ F_t: M \rightarrow N, $$ where $F_t(p) = F(t,p)$. $F$ is a time-dependent immersion or flow if for each $t \in [0,T)$, $F_t$ is an immersion. We will assume this below.

For each $t \in [0,T)$, $g_t = F_t^*h$ is a smooth $1$-parameter family of Riemannian metrics on $M$ and in particular, a section of $S^T_*M$. It can therefore be differentiated with respect to $t$, which results in a symmetric $2$-tensor, which is usually denoted $\partial_tg_t$ or $\partial_tg$ or $\dot{g}_t$.

You can also differentiate $F_t$ with respect to $t$. By definition, you get a map $$ \partial_tF_t: [0,T) \times M \rightarrow T_*N,$$ where for each $p \in M$ $$ \partial_tF_t(p) = c_p'(t), $$ where $c_p: [0,T) \rightarrow N$ is the parameterized curve given by $$ c_p(t) = F_t(p). $$

Since $F_t$ is an immersion, we can define for each $t$ the unit normal $\nu_t$ and second fundamental form $\II_t$.

The most studied geometric flows are of the following form: Given an abstract vector space $V$, consider a function $\phi: S^2_+V^* \times S^2V^* \rightarrow \mathbb{R}$, where $S^2_+V^*$ denotes the space of inner products on $V$, consider the PDE $$ \partial_tF_t = \phi(g_t,\II_t)\nu_t. $$

For example, the mean curvature flow is when $$ \phi(g, s) = g^{ij}s_{ij} $$ and the Gauss curvature flow is when $$ \phi(g,s) = {\det}_g(s). $$

Observe that for each $p \in M$, each side of the equation is a vector in $T_{F_t(p)}N$. Technically, each side is a section of $F_t^*T_*N$, but this abstract observation is more confusing than helpful. However, it is important, because it means you have to differentiate with respect to $t$ carefully. For example, if $v, w \in T_pM$, $$ \partial_tg_t(v,w) = \partial_t(g_N(F_t(p))((F_t)_*v,(F_t)_*w))), $$ so the right side has three terms. Here, I recommend that you initially do everything with respect to local coordinates $x = (x^1, \dots, x^{n-1})$ on $M$ and $y = (y^1, \dots, y^n)$ on $N$ and view each $y^i$ as a function of both $t$ and $x$. The computations become more obvious if you do that.