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Let $\Lambda$ be a lattice in $\mathbb{R}^n$. For $\bar{x} \in \mathbb{R}^n$, let $\| \bar{x} \| = max_{1 \leq i \leq n} \{ |x_i| \}$, i.e. the sup norm. Let $\lambda_1, ..., \lambda_n$ be a successive minima of $\Lambda$ with respect to the sup norm.

I am interested in counting number of points $N(U) = \# \{ \bar{x} \in \Lambda : \| \bar{x} \| < U \}$ when say $\lambda_{j} \leq U < \lambda_{j+1}$. I am expecting some like $$ \frac{U^j}{\lambda_1 ... \lambda_j} \ll N(U) \ll \frac{U^{j}}{\lambda_1 ... \lambda_{j}} $$ when $\lambda_{j} \leq U < \lambda_{j+1}$, where the implicit constants may depend on $\det (\Lambda)$ and $n$, but not on $\Lambda$ itself.

I would appreciate any assistance, comments, or references. Thank you very much!

PS And I mean the inequalities as they are. Thank you!

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    $\begingroup$ Since you were asking about the $j$-volume of the parallelopiped $\mathcal P$ spanned by $v_1,\ldots,v_j$ in $\mathbb{R}^n$, you might find the following formula interesting. Let $A$ be the $n$-by-$j$ matrix the contains the vectors $v_1,\ldots,v_j$ as the columns. Let $B_1,B_2,\ldots,B_k$ denote all of the $j$-by-$j$ minors of $A$, so $k=\binom{n}{j}$. Then $\operatorname{Vol}(\mathcal P)^2 = \sum_{i=1}^k \det(B_i)^2$. $\endgroup$
    – Joe Silverman
    Commented Apr 7, 2015 at 18:11

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Your expectation is correct. Let $\lambda_{j} \leq U < \lambda_{j+1}$. There are independent lattice vectors $v_1,\dots,v_j\in\Lambda$ such that $\|v_i\|=\lambda_i$ for $1\leq i\leq j$.

Consider the linear combinations $ c_1v_1+\dots +c_j v_j\in\Lambda$ with integral coefficients $c_i\in\mathbb{Z}$ satisfying $|c_i|\leq U/(2n\lambda_i)$. Their sup-norms are $\leq U/2<U$, and their number is $\gg_n U^j/(\lambda_1\dots\lambda_j)$. This proves the lower bound.

Now consider the subspace $V_j=\mathbb{R}v_1+\dots+\mathbb{R}v_j$ equipped with the $j$-volume coming from the euclidean norm in $\mathbb{R}^n$, i.e. an orthonormal basis in $V_j$ spans a $j$-cube of $j$-volume $1$. Then, $\Lambda_j=\Lambda\cap V_j$ is a lattice in $V_j$ with successive minima $\lambda_1,\dots,\lambda_j$, hence by Minkowski's second theorem its fundamental parallelepiped $P_j$ has $j$-volume $$\text{vol}_j(P_j)\geq 2^{-j}\cdot\lambda_1\dots\lambda_j\cdot\text{vol}_j(B_j),$$ where $B_j=\{ \bar{x} \in V_j : \| \bar{x} \| \leq 1 \}.$ The sup-norm is majorized by the euclidean norm, hence $B_j$ contains a euclidean $j$-ball of radius $1$, which shows that $\text{vol}_j(P_j)\gg_j \lambda_1\dots\lambda_j$. On the other hand, $$\{ \bar{x} \in \Lambda : \| \bar{x} \| < U \}=\{ \bar{x} \in \Lambda_j : \| \bar{x} \| < U \},$$ whence the translates of $P_j$ by the elements of this set are pairwise disjoint and they all lie in $\{ \bar{x} \in V_j : \| \bar{x} \| < (j+1)U \}$ whose $j$-volume is $\ll_j U^j$. This proves the upper bound.

Remark. The above argument was enhanced and corrected by the OP's comments, for which I am grateful.

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    $\begingroup$ @SJY: You don't need any reference. The vectors $v_i$ for $1\leq i\leq j$ are linearly independent, hence they generate a full rank lattice in the vector space $V$. The $j$-volume simply means the volume on $V$ induced by the Euclidean distance on $\mathbb{R}^n$. Perhaps you might keep in mind that the sup-norm and the Euclidean norm are equivalent on $\mathbb{R}^n$, namely $\|v\|_\infty\leq\|v\|_2\leq \sqrt{n}\|v\|_\infty$. Good books on the geometry of numbers is the one by Gruber-Lekkerkerker and the one by Cassels. If you like my answer, accept it officially so that it turns green. $\endgroup$
    – GH from MO
    Commented Apr 4, 2015 at 4:01
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    $\begingroup$ You don't have to calculate the $j$-volume of the fundamental parallelepiped for the above proof. However, if you want to calculate it, then simply orthogonalize the basis $\{v_1,\dots,v_j\}$ to an orthogonal basis $\{w_1,\dots,w_j\}$, and then the $j$-volume in question is $\prod_i|w_i|$. Alternately, express each $v_i$ $(1\leq i\leq j)$ in an orthonormal basis, and form a $j\times j$ determinant out of the $j$ coefficient vectors. The $j$-volume is the absolute value of this determinant. $\endgroup$
    – GH from MO
    Commented Apr 5, 2015 at 3:42
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    $\begingroup$ Note that this is just how one calculates the volume of the fundamental parallelepiped in $\mathbb{R}^n$, since $V$ is isomorphic to $\mathbb{R}^j$. A $j$-dimensional subspace of $\mathbb{R}^n$ is just a copy of $\mathbb{R}^j$. $\endgroup$
    – GH from MO
    Commented Apr 5, 2015 at 3:45
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    $\begingroup$ In my previous comment $|w_i|$ denotes the Euclidean norm of $w_i$, not its sup-norm. $\endgroup$
    – GH from MO
    Commented Apr 5, 2015 at 13:43
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    $\begingroup$ The set $\{\bar{x} \in \Gamma : \| \bar{x} \| <U\}$ is the same as $\{\bar{x} \in \Lambda_j : \| \bar{x} \| <U\}$, since $U<\lambda_{j+1}$. Hence the translates of $P_j$ by the elements of this set are pairwise disjoint (by the definition of $P_j$). $\endgroup$
    – GH from MO
    Commented Apr 7, 2015 at 8:55

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