Fiber surfaces in three-manifolds minimize genus

Here's a cool fact.  If you have a fibered three-manifold, i.e. your three-manifold is a fiber bundle with fiber a surface and base $S^1$, then the fiber surface minimizes genus among all surfaces in its homology class.  (If the fibered three-manifold has boundary, necessarily a union of tori, then we think about the homology class of the surface rel boundary.)   This says that for fibered knots in $S^3$, the fiber surface is a minimal genus Seifert surface!  Just for funsies I wanted to put a proof of this here, because I think it's a nice idea that shows up in various guises in low-dimensional topology.  

If you haven't thought much about fibered three-manifolds, some good examples to keep in mind are: $\Sigma^2 \times S^1$, the complement of a torus knot (here is an awesome visualization), the complement of the figure eight knot, or 0-surgery on any fibered knot in the three-sphere.      

Let $Y$ be a fibered three-manifold with a fiber $\Sigma$.  Form the infinite cyclic cover of $Y$, denoted $\widetilde{Y}$, by cutting $Y$ along $\Sigma$ and gluing infinitely many copies together end to end.  (A fancier way of constructing this is that the Poincare dual of $\Sigma$ is an element of $H^1(Y)$, i.e. a map from $\pi_1(Y)$ to $\mathbb{Z}$, and this cover is the one corresponding to the kernel of this map.)  Since $\Sigma$ is a fiber, $Y - \Sigma$ is just $\Sigma \times I$ and so $\widetilde{Y}$ is $\Sigma \times \mathbb{R}$.  Let $\Sigma'$ be any surface in $Y$ which is homologous to $\Sigma$.  Then, you can check that $\Sigma'$ actually lifts to the cover $\widetilde{Y}$ and in the cover $\Sigma$ and $\Sigma'$ are homologous.  (Here's one perspective, loosely sketched.  Since $[\Sigma] = [\Sigma']$, their Poincare duals are the same, so the covers associated to each are the same.  But the latter cover is the one where we cut along $\Sigma'$ and stack together infinitely many copies, which $\Sigma'$ certainly lifts to.)  Now, define a map from $\Sigma'$ to $\Sigma$ as follows: include $\Sigma'$ into $\Sigma \times \mathbb{R}$ and project onto $\Sigma$; because $\Sigma'$ and $\Sigma$ are homologous in $\Sigma \times \mathbb{R}$, this map induces an isomorphism on $H_2$.  Get excited because that means this is a degree one map from $\Sigma'$ to $\Sigma$, and so that means $b_1(\Sigma') \geq b_1(\Sigma)$!  (See the previous post on degree one maps.)  In other words, $g(\Sigma') \geq g(\Sigma)$.  

This idea is pretty useful in other settings.  Here's an example of a well-known application to Floer homology.  Let $Y$ be a three-manifold and suppose that $Z$ is a homology sphere that embeds in $Y \times S^1$.  Then, $Y$ is a homology sphere and $\dim F(Z) \geq \dim F(Y)$, where $F$ is basically any Floer homology you can think of.  As before, build the associated infinite cyclic cover of $Y \times S^1$ which is $Y \times \mathbb{R}$.  Again, $Z$ will lift and thus embeds in $Y \times \mathbb{R}$ such that $Z$ and $Y$ are homologous (as elements of $H_3$), and we can get a degree one map from $Z$ to $Y$ which means $\pi_1(Z)$ surjects $\pi_1(Y)$.  Since $H_1(Z) = 0$, this means that $H_1(Y) = 0$ as well, so $Y$ is a homology three-sphere.  Now, for the Floer homology part.  By compactness, let's really think of $Z$ as embedding in $Y \times I$ inside $Y \times \mathbb{R}$; so we can think of $Y \times I$ as the union of a cobordism $W_1$ from $Y$ to $Z$ and another $W_2$ from $Z$ to $Y$.  Now, the map on Floer homology for $Y \times I$ is the identity on $F(Y)$.  But, the functoriality for cobordism maps in Floer homology says that $F(Y \times I) = F(W_2) \circ F(W_1)$.  That means the identity map has to factor through $F(Z)$.  This is only possible if $\dim F(Z) \geq \dim F(Y)$!