Collection of sub line bundles of a vector bundle

Let X/k be a projective integral variety over a field, with fixed $\mathcal{O}(1)$, $\mathcal{E}$ be a rank two vector bundle, for simplicity assume it is stable. Regarding the collection of sub line bundles $\mathcal{L}\subset \mathcal{E}$ up to isomorphism,

EDIT: and moreover $\mathcal{L}$ is the line bundle given by an effective divisor,

denoted this collection by $Sub(\mathcal{E})$, I have the following questions:

(1) When is this collection finite?

(2) Take another scheme $T/k$, and base change $X_T/T$, when is $Sub(\mathcal{E}_T)=Sub(\mathcal{E})_T$ ?

(3) If the above equality doesn't hold, when does $Sub(\mathcal{E}_T)=Sub(\mathcal{E})_T$ up to tensoring pull back of a line bundle on T?

(4) If the answer to (2) and (3) are not very trivial, then consider the functor $Sub(\mathcal{E}_T\otimes f_T^*(\mathcal{N}))$, where $f:X\rightarrow \mathcal{S}pec\ k$ is the structure morphism, $\mathcal{N}$ is a line bundle on T, and we identify two sub line bundles of $\mathcal{E}$ up to tensoring with $f^*_T(\mathcal{N})$. When is this functor representable?

Edit: Here by a sub line bundle I just mean a rank one subsheaf which is itself a line bundle; i.e. in an exact sequence like $0\rightarrow \mathcal{L}\rightarrow \mathcal{E}\rightarrow \mathcal{Q}\rightarrow 0$, I didn't mean both $\mathcal{L}$ and the "quotient" $\mathcal{Q}$ are line bundles.

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Could someone please tell me what is wrong with the formatting? It is horrible, but I don't know what went wrong... –  Ying Zhang Oct 18 '11 at 4:07
I fixed the formatting. Sometimes you need to add backticks, as explained in "How to write math" on this page. –  Angelo Oct 18 '11 at 4:10
Consider a vector bundle on $\mathbb P^1$, say the rank-2 trivial bundle. I argue that it contains sub-line bundles of arbitrarily low degree. The section $(x^n,1)$ has a degree $n$ pole at $\infty$ but no zeroes. Furthermore it generates a line bundle, which will have degree $-n$. For this reason, I suspect the collection is rarely finite. What is an effective line bundle? Is it a line bundle given by an effective divisor? On $\mathbb P^1$, at least, those are finite, because there is a maximum degree. Similar results should hold over, at least, every curve. –  Will Sawin Oct 18 '11 at 4:42
It is certainly true that for any curve the collection is always infinite; on the other hand, it is easy to give examples in higher dimension where it is empty. I doubt that there is a general criterion. As a general comment, questions of the type "When does this happen" are rather hard to answer. The OP should really make an effort to explain what he has in mind –  Angelo Oct 18 '11 at 4:58
If one has the property that $H^1(X,L)=0$ for every line bundle $L$ on $X$, then this is equivalent to saying $Ext^1(L_1,L_2)=0$ for any line bundles $L_1,L_2$. In this setting, if a rank $2$ vector bundle $E$ admits a line subbundle $L$, then the short exact sequence $0 \rightarrow L \rightarrow E \rightarrow Q \rightarrow 0$ must split, so $E$ has only $2$ possible line subbundles. –  Parsa Oct 18 '11 at 6:08