Thank you, this is a nice example. I got fooled by my commutative heritage.

I believe that every nilpotent subspace $V$ can be conjugated inside the space of uppertriangular matrices, regardless of your assumption on dimensions of $V^k$. So they will have a common invariant zero eigenspace.

First of all, my example shows that if you take two generic uppertriangular matrices $A$ and $B$ of size $9$, then $A^2,AB,B^2$ are linearly independent. I suspect that in fact $BA$ would also be independent from them in this case, but I did not check it. I also think that taking $V$ to be a generic dimension two subspace of uppertriangular size $4$ matrices, would give $V^2$ of dimension $3$ (elements of $V^2$ can only have three possible nonzero entries, but I see no reason for why the span of them would be any less than that).

@WillSawin Great, I can definitely follow it now. Yes, I think this calculation would indicate that the restriction of this line bundle to $X_0(n)$ would only be the sum of the divisors obtained from traces of Heegner points, so it would not be any "new" bundle.

@AbhinavKumar No, can you please send me a reference? This is more or less just a hobby for me at this point, so my knowledge of the literature is very spotty.

@AbhinavKumar Indeed, theorem 7.6 of Dolgachev's paper gives some other description of the corresponding K3. Perhaps, it is more interesting than the Kummer K3, although either one may be suitable for the purposes of finding interesting points on $X_0(n)$ over $\bar{\mathbb Q}$.

I am not interested in rational points on $X_0(n)^+$ as much as I am interested in rational divisors on it. Basically, I am after a more advanced version of the Heegner points -- some interesting points on $X_0(n)$ defined over $\bar {\mathbb Q}$ such that their images on elliptic curves of rank higher than $1$ give nontrivial traces. Note that Heegner points would only give traces that are torsion by Gross-Zagier formula.

I don't think that these K3-s are all that hard to construct from the elliptic curves. My best guess is that if $E_1\to E_2$ is the corresponding cyclic isogeny, then the K3 is the resolution of $E_1\times E_2/<(-1,-1)>$, but I have not tried to verify it.

Right. I am somehow hoping that by looking at a more complicated object, namely the associated K3 surface, one might discover something that is not visible at the elliptic curve level.

Sorry, just glossed over that requirement in the problem setting.

Well, $f(x)=\cos(x)$ certainly would give linearly dependent $f(x+\alpha)$. So the assumptions needed to apply Fourier transform are relevant.

Will, I was happy with your answer, thank you! I didn't quite follow the arguments regarding the bottom corner, but, certainly, if your Heegner point guess is true, then it would dash my faint hope of getting some more sporadic divisors from this construction. I assume that these would be Heegner points for the conductor not coprime (equal?) to the level, which are generally avoided in applications like Gross-Zagier formula. Am I correct?