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taylor
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I am able to prove the followsfollowing two propositions: Recall that the $i$-th successive minimum of $L\in \mathcal L$, denoted $\lambda_i(L)$ is the infimum of the radii of the balls containing $i$-linearly independent vectors in $L$. All norms are Euclidean 2-norm.

Let $\Lambda$ be a lattice $\mathbb R^d$. Assume that $d\le 4$, then there exist a basis $\{v_1,\dots,v_d\}$ of $\Lambda$ such that $$\|v_j\|=\lambda_j(\Lambda), ~\text{for}~ j=\{1,2,\dots,d\}.$$

and

Let $\Lambda$ be a lattice in $\mathbb R^d$. Then there exist a basis $v_1,v_2,\dots,v_d$ of $\Lambda$ such that $$\|v_1\|=\lambda_1(\Lambda),\|v_2\|_d \asymp_d \lambda_2(\Lambda),\dots,\|v_d\| \asymp_d \lambda_d(\Lambda).$$

But I am very puzzled if the follow combination of the above two is true:

Let $\Lambda$ be a lattice in $\mathbb R^d$. Then there exist a basis $v_1,v_2,\dots,v_d$ of $\Lambda$ such that $$\|v_i\|=\lambda_i(\Lambda) (1\le i \le 4),\|v_5\|_d \asymp_d \lambda_5(\Lambda),\dots,\|v_d\| \asymp_d \lambda_d(\Lambda).$$

Is the last statement correct? I don't see how it follows from the previous two statements directly. There is some magic with the dimension 4. The example $\text{Span}\{e_1,e_2,\dots, e_d, \frac{1}{2}(e_1+\dots+e_d)\}$ provides the obstruction for the reduction process for $d\ge 5$.

I am able to prove the follows propositions: Recall that the $i$-th successive minimum of $L\in \mathcal L$, denoted $\lambda_i(L)$ is the infimum of the radii of the balls containing $i$-linearly independent vectors in $L$. All norms are Euclidean 2-norm.

Let $\Lambda$ be a lattice $\mathbb R^d$. Assume that $d\le 4$, then there exist a basis $\{v_1,\dots,v_d\}$ of $\Lambda$ such that $$\|v_j\|=\lambda_j(\Lambda), ~\text{for}~ j=\{1,2,\dots,d\}.$$

and

Let $\Lambda$ be a lattice in $\mathbb R^d$. Then there exist a basis $v_1,v_2,\dots,v_d$ of $\Lambda$ such that $$\|v_1\|=\lambda_1(\Lambda),\|v_2\|_d \asymp_d \lambda_2(\Lambda),\dots,\|v_d\| \asymp_d \lambda_d(\Lambda).$$

But I am very puzzled if the follow combination of the above two is true:

Let $\Lambda$ be a lattice in $\mathbb R^d$. Then there exist a basis $v_1,v_2,\dots,v_d$ of $\Lambda$ such that $$\|v_i\|=\lambda_i(\Lambda) (1\le i \le 4),\|v_5\|_d \asymp_d \lambda_5(\Lambda),\dots,\|v_d\| \asymp_d \lambda_d(\Lambda).$$

Is the last statement correct? I don't see how it follows from the previous two statements directly. There is some magic with the dimension 4. The example $\text{Span}\{e_1,e_2,\dots, e_d, \frac{1}{2}(e_1+\dots+e_d)\}$ provides the obstruction for the reduction process for $d\ge 5$.

I am able to prove the following two propositions: Recall that the $i$-th successive minimum of $L\in \mathcal L$, denoted $\lambda_i(L)$ is the infimum of the radii of the balls containing $i$-linearly independent vectors in $L$. All norms are Euclidean 2-norm.

Let $\Lambda$ be a lattice $\mathbb R^d$. Assume that $d\le 4$, then there exist a basis $\{v_1,\dots,v_d\}$ of $\Lambda$ such that $$\|v_j\|=\lambda_j(\Lambda), ~\text{for}~ j=\{1,2,\dots,d\}.$$

and

Let $\Lambda$ be a lattice in $\mathbb R^d$. Then there exist a basis $v_1,v_2,\dots,v_d$ of $\Lambda$ such that $$\|v_1\|=\lambda_1(\Lambda),\|v_2\|_d \asymp_d \lambda_2(\Lambda),\dots,\|v_d\| \asymp_d \lambda_d(\Lambda).$$

But I am very puzzled if the follow combination of the above two is true:

Let $\Lambda$ be a lattice in $\mathbb R^d$. Then there exist a basis $v_1,v_2,\dots,v_d$ of $\Lambda$ such that $$\|v_i\|=\lambda_i(\Lambda) (1\le i \le 4),\|v_5\|_d \asymp_d \lambda_5(\Lambda),\dots,\|v_d\| \asymp_d \lambda_d(\Lambda).$$

Is the last statement correct? I don't see how it follows from the previous two statements directly. There is some magic with the dimension 4. The example $\text{Span}\{e_1,e_2,\dots, e_d, \frac{1}{2}(e_1+\dots+e_d)\}$ provides the obstruction for the reduction process for $d\ge 5$.

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taylor
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I am able to prove the follows propositions: Recall that the $i$-th successive minimum of $L\in \mathcal L$, denoted $\lambda_i(L)$ is the infimum of the radii of the balls containing $i$-linearly independent vectors in $L$. All norms are Euclidean 2-norm.

Let $\Lambda$ be a lattice $\mathbb R^d$. Assume that $d\le 4$, then there exist a basis $\{v_1,\dots,v_d\}$ of $\Lambda$ such that $$\|v_j\|=\lambda_j, ~\text{for}~ j=\{1,2,\dots,d\}.$$$$\|v_j\|=\lambda_j(\Lambda), ~\text{for}~ j=\{1,2,\dots,d\}.$$

and

Let $\Lambda$ be a lattice in $\mathbb R^d$. Then there exist a basis $v_1,v_2,\dots,v_d$ of $\Lambda$ such that $$\|v_1\|=\lambda_1(\Lambda),\|v_2\|_d \asymp_d \lambda_2(\Lambda),\dots,\|v_d\| \asymp_d \lambda_d(\Lambda).$$

But I am very puzzled if the follow combination of the above two is true:

Let $\Lambda$ be a lattice in $\mathbb R^d$. Then there exist a basis $v_1,v_2,\dots,v_d$ of $\Lambda$ such that $$\|v_1\|=\lambda_1(\Lambda),\|v_2\|_d \asymp_d \lambda_2(\Lambda),\dots,\|v_d\| \asymp_d \lambda_d(\Lambda).$$ and $v_1,v_2,v_3,v_4$ can be chosen such that their lengths are precisely equal to $\lambda_i(\Lambda)$ for $i=1,2,3,4$.$$\|v_i\|=\lambda_i(\Lambda) (1\le i \le 4),\|v_5\|_d \asymp_d \lambda_5(\Lambda),\dots,\|v_d\| \asymp_d \lambda_d(\Lambda).$$

Is the last statement correct? I don't see how it follows from the previous two statements directly. There is some magic with the dimension 4. The example $\text{Span}\{e_1,e_2,\dots, e_d, \frac{1}{2}(e_1+\dots+e_d)\}$ provides the obstruction for the reduction process for $d\ge 5$.

I am able to prove the follows propositions: Recall that the $i$-th successive minimum of $L\in \mathcal L$, denoted $\lambda_i(L)$ is the infimum of the radii of the balls containing $i$-linearly independent vectors in $L$. All norms are Euclidean 2-norm.

Let $\Lambda$ be a lattice $\mathbb R^d$. Assume that $d\le 4$, then there exist a basis $\{v_1,\dots,v_d\}$ of $\Lambda$ such that $$\|v_j\|=\lambda_j, ~\text{for}~ j=\{1,2,\dots,d\}.$$

and

Let $\Lambda$ be a lattice in $\mathbb R^d$. Then there exist a basis $v_1,v_2,\dots,v_d$ of $\Lambda$ such that $$\|v_1\|=\lambda_1(\Lambda),\|v_2\|_d \asymp_d \lambda_2(\Lambda),\dots,\|v_d\| \asymp_d \lambda_d(\Lambda).$$

But I am very puzzled if the follow combination of the above two is true:

Let $\Lambda$ be a lattice in $\mathbb R^d$. Then there exist a basis $v_1,v_2,\dots,v_d$ of $\Lambda$ such that $$\|v_1\|=\lambda_1(\Lambda),\|v_2\|_d \asymp_d \lambda_2(\Lambda),\dots,\|v_d\| \asymp_d \lambda_d(\Lambda).$$ and $v_1,v_2,v_3,v_4$ can be chosen such that their lengths are precisely equal to $\lambda_i(\Lambda)$ for $i=1,2,3,4$.

Is the last statement correct? I don't see how it follows from the previous two statements directly. There is some magic with the dimension 4. The example $\text{Span}\{e_1,e_2,\dots, e_d, \frac{1}{2}(e_1+\dots+e_d)\}$ provides the obstruction for the reduction process for $d\ge 5$.

I am able to prove the follows propositions: Recall that the $i$-th successive minimum of $L\in \mathcal L$, denoted $\lambda_i(L)$ is the infimum of the radii of the balls containing $i$-linearly independent vectors in $L$. All norms are Euclidean 2-norm.

Let $\Lambda$ be a lattice $\mathbb R^d$. Assume that $d\le 4$, then there exist a basis $\{v_1,\dots,v_d\}$ of $\Lambda$ such that $$\|v_j\|=\lambda_j(\Lambda), ~\text{for}~ j=\{1,2,\dots,d\}.$$

and

Let $\Lambda$ be a lattice in $\mathbb R^d$. Then there exist a basis $v_1,v_2,\dots,v_d$ of $\Lambda$ such that $$\|v_1\|=\lambda_1(\Lambda),\|v_2\|_d \asymp_d \lambda_2(\Lambda),\dots,\|v_d\| \asymp_d \lambda_d(\Lambda).$$

But I am very puzzled if the follow combination of the above two is true:

Let $\Lambda$ be a lattice in $\mathbb R^d$. Then there exist a basis $v_1,v_2,\dots,v_d$ of $\Lambda$ such that $$\|v_i\|=\lambda_i(\Lambda) (1\le i \le 4),\|v_5\|_d \asymp_d \lambda_5(\Lambda),\dots,\|v_d\| \asymp_d \lambda_d(\Lambda).$$

Is the last statement correct? I don't see how it follows from the previous two statements directly. There is some magic with the dimension 4. The example $\text{Span}\{e_1,e_2,\dots, e_d, \frac{1}{2}(e_1+\dots+e_d)\}$ provides the obstruction for the reduction process for $d\ge 5$.

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taylor
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