I'm trying to relate the slope decomposition of a product of linear operators to the slope decompositions with regard to each of the operators in the product. 


First I'll give some background, for which I'm following section 2.3 of Urban's Eigenvarietes for reductive groups.


If $L/\mathbb{Q}\_p$ is a finite extension. say a polynomial $P(X)$ of degree $d$ has slope $\leq h$ if $P(0)$ is a unit in the ring of integers of $L$ and the roots of $P^\*(X)=X^dP(1/X)$ in $\bar{\mathbb{Q}}\_p$ have $p$-adic valuation less than $h$. 


Then if $M$ is a vector space over $L$ and $U$ is a continuous linear operator $M$, we say that $M$ has a slope $\leq h$ decomposition with respect to $U$, if we can write $M:=M\_1 \oplus M\_2$, where both $M\_1,M\_2$ are stable under the action of $U$ and we have:


(1) $M\_1$ is finite dimensional over $L$.


(2) The polynomial $\det(1-X\cdot U)|M\_1$ is of slope $\leq h$.


(3) For any polynomial $P$ of slope $\leq h$, the restriction of $P^\*(U)$ to $M\_2$ is an invertible endomorphism of $M\_2$.


Now my question is that if I have $U=\prod\_{i=1}^{n} U\_i$, where $U,U\_i$ are all linear operators on $M$. Then if I take an element that has slope $\leq h$ with respect to $U$, does it follow that it will have slope $\leq h$ with respect to all the $U\_i$.


Or more generally is there a relation between the slope decomposition of with respect to $U$ and the slope decomposition with respect to the $U\_i$.


Note: For the case I'm looking at I can actually take the operators to be compact operators, and my $M$ will be spaces of modular forms. But I don't know how much this helps.



Slope decomposition of a product of operators
The answer to your exact question is clearly "no" even if $M$ is finite-dimensional. Any operator on a finite-dimensional space has a slope decomposition, and if $n = 2$ and $U\_2 = U\_1^{-1}$ then every element has slope $0$ for $U = U\_1 U\_2 = \operatorname{id}$, but that doesn't force every element to have slope $0$ for the $U\_i$.


To restore some sanity, it might be best to assume that 


* $M$ is a Banach space,
* the $U\_i$ are compact,
* every $U\_i$ has operator norm $\le 1$,
* the $U\_i$ commute with each other (this is really the key point).


Then $M$ has a "slope $\le (h\_1, \dots, h\_n)$ decomposition for $(U\_1, \dots, U\_n)$" for any $h\_i \ge 0$; and there are easy containments
$$M^{\le h} \subseteq M^{\le (h\_1, \dots, h\_n)} \subseteq M^{\le H} $$


where $H = \sum h\_i$ and $h = \min h\_i$. But both of these will be strict in general.


It's probably best to think about this in the case $n = 2$, where you can visualize the slopes of simultaneous eigenspaces (or generalized eigenspaces) for $U\_1$ and $U\_2$ as points in $(\mathbf{R}\_{\ge 0})^2$, and imposing a small-slope condition for $U\_1$, $U\_2$, or $U$ amounts to chopping off at a horizontal, vertical, or diagonal line. 


(Compare Emerton's "Jacquet modules I" paper, where he has a commutative monoid of operators acting, corresponding to some torus in a reductive group, and he defines slopes as living in the character group of that torus tensored with $\mathbf{R}$ -- I suspect this may be somewhat related to the examples you have in mind.)


