Let $f:X\to S$ be a "nice" morphism of "nice" schemes. Let $L$ be an ample line bundle on $X$.

When is $\det f_\ast L$ also ample?

A "nice" morphism could be anything from "finite type separated" to "flat projective" or "birational proper surjective".

A "nice" scheme could be "integral non-singular", or "integral normal", or "with easy singularities".

  • 2
    $\begingroup$ One of the common situations which arise is the following. Assume $f$ is flat and projective and let $L=f^*A\otimes B$ where $A$ is ample on $S$. Further assume that $f_*B$ is globally generated and higher direct images vanish. Then $f_*B$ is a vector bundle and $f_*L=A\otimes f_*B$ is ample and thus, so is its determinant. So, typically, if $A$ is ample on $S$, $B$ is $f-ample on $X$, then for all large $n$, $f^*A\otimes B^n$ will have this property. $\endgroup$
    – Mohan
    Apr 12, 2013 at 16:55

2 Answers 2


Consider a morphism $f:\mathbb{P}^1 \to \mathbb{P}^1$ of degree $e> 2$. Using Grothendieck's splitting lemma, it is straightforward to compute that $f_*\mathcal{O}(1)$ is isomorphic to $\mathcal{O}^{\oplus 2} \oplus \mathcal{O}(-1)^{\oplus (e-2)}$. Thus the determinant is $\mathcal{O}(-(e-2))$, which is anti-ample. Thus, I don't see any reason at all that $\text{det}(f_*L)$ should be ample in general.


Actually, I think that the phenomenon Jason is describing is very much due to using a finite morphism, so there is something intelligent one could say about this.

Of course, as stated there is no chance that simply assuming something nice about $f$ would induce the desired result, because $f_*L$ could easily be zero. So one needs to assume something about $L$ as well.

So we probably need something like $L$ is "sufficiently ample". Once we allow that then experience shows that it is usually more tractable if we can compare our bundle to the canonical bundle. In other words, let us write $L=\omega_{X/S}\otimes M$ for some $\mathbb Q$-line bundle $M$ (so in particular assume that $\omega_{X/S}$ is a $\mathbb Q$-line bundle).

For simplicity let us assume that $M$ is actually a line bundle, which is the case if $f$ is a Gorenstein morphism, so in particular if $X$ and $S$ are non-singular.

Now if $M$ is very ample (or in characteristic zero generated by global sections) then Let $D$ be the zero locus of a general section of $M$. This will not be much more singular than $X$ is, for instance if $X$ is smooth, then so is $D$. If, say, $X$ has no worse than canonical singularities, then $(X,D)$ is log canonical and by the assumption that $L$ is ample and assuming that $f$ is flat with connected fibers it follows that $f$ is a family of stable log varieties and it should also follow that it has maximal variation in moduli.

Under these conditions it follows that $\det (f_*L^q)$ is big by 7.1 of arxiv:1503.02952. Since the conditions survive after restricting to any closed subvariety of the base, this implies that it is actually ample.

One could make this a little more general by allowing more singularities, but the spirit would remain the same. I suppose this might seem far from what you asked for, but I doubt that one can have a much more general statement than this. One could say that since there are many theorems that say that certain push-forwards are ample or big, this is likely a hard problem, although, truth be told, those theorems usually do not assume that $L$ is ample, but, say that it is $f$-ample. In that case without additional assumption, such as a relation to the relative canonical sheaf, there is clearly no chance for anything like this to be true (think if Mohan's example and make $B$ very negative).


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