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I think you probably can decide this without tears, but just because your matrices are positive definite. Notice that, if $x$ is a row vector of integers, $$ x A x^T = x Q B Q^T x^T = y B y^T, $$ where we take $y = x Q.$ So, your two matrices have the same determinants, but they also have the same $\mathbb Z$-minimum. Indeed, I think they have the same "successive minima," and I would not be surprised if that were sufficient as well. In any case, you ought to get Gruber and Lekkerkerker. Note that emaphasis is switched from the matrix $A$ to the compact set with nonempty interior, $$ x A x^T \leq 1. $$

Now, why is is possible to find the $\mathbb Z$-minimum, which is some peculiar real number in your setting? Begin with $ m $ being the smallest diagonal entry of $A.$ The set of integer vectors $x$ with $ x A x^T \leq m$ is bounded by eigenvalue considerations (I actually do a little thing with Lagrange multipliers to get a separate bound for each coordinate $x_i.$) This gives a cube or rectangular box depending how you do it, leading to a finite set of integer vectors. So finding the $\mathbb Z$-minimum is a finite check.

I have seen this problem in infinite detail only for indefinite binary forms, see Cusick and Flahive

Alright, I am remembering things. Minkowski reduction is not definitive even for integral quadratic forms for dimensions at least three. It is quite good in dimensions 3,4. Alexander Schiemman made an improvement to Eisenstein's reduction in dimension 3, and wrote it up allowing for real entries, Mathematische Annalen, volume 308 (1997) pages 507-517, Ternary Positive Definite Quadratic Forms are determined by their Theta Series.

EDIT, 3:52 pm Pacific: Some of Schiemann's favorite themes give the simplest view of this. By definiteness, you can make two lists of represented values up to some bound, where "represented" means by an integral vector $x$ as above. Although these are real numbers, let me still call each list a "theta series," because we are able to count the number of times each real number is represented. Now, it is necessary that the two matrices have the same "theta series" in order to be $\mathbb Z$-equivalent. However, if the dimension is 4 or larger, this is not sufficient. The original examples of disagreement (same theta series but not equivalent) were by Milnor, dimension 16. The general topic is loosely under the heading of "Can you hear the shape of a drum?" Anyway, there was progress lowering the dimension, especially by Kneser and then Kitaoka. Finally, Schiemann found a four-dimensional pair, Arch. Math. 54 (1990) pages 372-375. Conway was able to include a version in his book The Sensual Quadratic Form. A less Conwayish description, including just coefficients, in in Quaternary Quadratic Forms by Gordon L. Nipp. There we go, page 1 and then again on page 110, in genus 4: the final two forms with minimum 2, after which there are three forms with minimum 3 and that is the end of that genus and that discriminiminiminant.

I think you probably can decide this without tears, but just because your matrices are positive definite. Notice that, if $x$ is a row vector of integers, $$ x A x^T = x Q B Q^T x^T = y B y^T, $$ where we take $y = x Q.$ So, your two matrices have the same determinants, but they also have the same $\mathbb Z$-minimum. Indeed, I think they have the same "successive minima," and I would not be surprised if that were sufficient as well. In any case, you ought to get Gruber and Lekkerkerker. Note that emaphasis is switched from the matrix $A$ to the compact set with nonempty interior, $$ x A x^T \leq 1. $$

Now, why is is possible to find the $\mathbb Z$-minimum, which is some peculiar real number in your setting? Begin with $ m $ being the smallest diagonal entry of $A.$ The set of integer vectors $x$ with $ x A x^T \leq m$ is bounded by eigenvalue considerations (I actually do a little thing with Lagrange multipliers to get a separate bound for each coordinate $x_i.$) This gives a cube or rectangular box depending how you do it, leading to a finite set of integer vectors. So finding the $\mathbb Z$-minimum is a finite check.

I have seen this problem in infinite detail only for indefinite binary forms, see Cusick and Flahive

Alright, I am remembering things. Minkowski reduction is not definitive even for integral quadratic forms for dimensions at least three. It is quite good in dimensions 3,4. Alexander Schiemman made an improvement to Eisenstein's reduction in dimension 3, and wrote it up allowing for real entries, Mathematische Annalen, volume 308 (1997) pages 507-517, Ternary Positive Definite Quadratic Forms are determined by their Theta Series.

EDIT, 3:52 pm Pacific: Some of Schiemann's favorite themes give the simplest view of this. By definiteness, you can make two lists of represented values up to some bound, where "represented" means by an integral vector $x$ as above. Although these are real numbers, let me still call each list a "theta series," because we are able to count the number of times each real number is represented. Now, it is necessary that the two matrices have the same "theta series" in order to be $\mathbb Z$-equivalent. However, if the dimension is 4 or larger, this is not sufficient. The original examples of disagreement (same theta series but not equivalent) were by Witt, dimension 16, later used by Milnor. The general topic is loosely under the heading of "Can you hear the shape of a drum?" Anyway, there was progress lowering the dimension, especially by Kneser and then Kitaoka. Finally, Schiemann found a four-dimensional pair, Arch. Math. 54 (1990) pages 372-375. Conway was able to include a version in his book The Sensual Quadratic Form. A less Conwayish description, including just coefficients, in in Quaternary Quadratic Forms by Gordon L. Nipp. There we go, page 1 and then again on page 110, in genus 4: the final two forms with minimum 2, after which there are three forms with minimum 3 and that is the end of that genus and that discriminiminiminant.

I think you probably can decide this without tears, but just because your matrices are positive definite. Notice that, if $x$ is a row vector of integers, $$ x A x^T = x Q B Q^T x^T = y B y^T, $$ where we take $y = x Q.$ So, your two matrices have the same determinants, but they also have the same $\mathbb Z$-minimum. Indeed, I think they have the same "successive minima," and I would not be surprised if that were sufficient as well. In any case, you ought to get Gruber and Lekkerkerker. Note that emaphasis is switched from the matrix $A$ to the compact set with nonempty interior, $$ x A x^T \leq 1. $$

Now, why is is possible to find the $\mathbb Z$-minimum, which is some peculiar real number in your setting? Begin with $ m $ being the smallest diagonal entry of $A.$ The set of integer vectors $x$ with $ x A x^T \leq m$ is bounded by eigenvalue considerations (I actually do a little thing with Lagrange multipliers to get a separate bound for each coordinate $x_i.$) This gives a cube or rectangular box depending how you do it, leading to a finite set of integer vectors. So finding the $\mathbb Z$-minimum is a finite check.

I have seen this problem in infinite detail only for indefinite binary forms, see Cusick and Flahive

Alright, I am remembering things. Minkowski reduction is not definitive even for integral quadratic forms for dimensions at least three. It is quite good in dimensions 3,4. Alexander Schiemman made an improvement to Eisenstein's reduction in dimension 3, and wrote it up allowing for real entries, Mathematische Annalen, volume 308 (1997) pages 507-517, Ternary Positive Definite Quadratic Forms are determined by their Theta Series.

EIT, 3:52 pm Pacific: Some of Schiemann's favorite themes give the simplest view of this. By definiteness, you can make two lists of represented values up to some bound, where "represented" means by an integral vector $x$ as above. Although these are real numbers, let me still call each list a "theta series," because we are able to count the number of times each real number is represented. Now, it is necessary that the two matrices have the same "theta series" in order to be $\mathbb Z$-equivalent. However, if the dimension is 4 or larger, this is not sufficient. The original examples of disagreement (same theta series but not equivalent) were by Milnor, dimension 16. The general topic is loosely under the heading of "Can you hear the shape of a drum?" Anyway, there was progress lowering the dimension, especially by Kneser and then Kitaoka. Finally, Schiemann found a four-dimensional pair, Arch. Math. 54 (1990) pages 372-375. Conway was able to include a version in his book The Sensual Quadratic Form. A less Conwayish description, including just coefficients, in in Quaternary Quadratic Forms by Gordon L. Nipp.

I think you probably can decide this without tears, but just because your matrices are positive definite. Notice that, if $x$ is a row vector of integers, $$ x A x^T = x Q B Q^T x^T = y B y^T, $$ where we take $y = x Q.$ So, your two matrices have the same determinants, but they also have the same $\mathbb Z$-minimum. Indeed, I think they have the same "successive minima," and I would not be surprised if that were sufficient as well. In any case, you ought to get Gruber and Lekkerkerker. Note that emaphasis is switched from the matrix $A$ to the compact set with nonempty interior, $$ x A x^T \leq 1. $$

Now, why is is possible to find the $\mathbb Z$-minimum, which is some peculiar real number in your setting? Begin with $ m $ being the smallest diagonal entry of $A.$ The set of integer vectors $x$ with $ x A x^T \leq m$ is bounded by eigenvalue considerations (I actually do a little thing with Lagrange multipliers to get a separate bound for each coordinate $x_i.$) This gives a cube or rectangular box depending how you do it, leading to a finite set of integer vectors. So finding the $\mathbb Z$-minimum is a finite check.

I have seen this problem in infinite detail only for indefinite binary forms, see Cusick and Flahive

Alright, I am remembering things. Minkowski reduction is not definitive even for integral quadratic forms for dimensions at least three. It is quite good in dimensions 3,4. Alexander Schiemman made an improvement to Eisenstein's reduction in dimension 3, and wrote it up allowing for real entries, Mathematische Annalen, volume 308 (1997) pages 507-517, Ternary Positive Definite Quadratic Forms are determined by their Theta Series.

EDIT, 3:52 pm Pacific: Some of Schiemann's favorite themes give the simplest view of this. By definiteness, you can make two lists of represented values up to some bound, where "represented" means by an integral vector $x$ as above. Although these are real numbers, let me still call each list a "theta series," because we are able to count the number of times each real number is represented. Now, it is necessary that the two matrices have the same "theta series" in order to be $\mathbb Z$-equivalent. However, if the dimension is 4 or larger, this is not sufficient. The original examples of disagreement (same theta series but not equivalent) were by Milnor, dimension 16. The general topic is loosely under the heading of "Can you hear the shape of a drum?" Anyway, there was progress lowering the dimension, especially by Kneser and then Kitaoka. Finally, Schiemann found a four-dimensional pair, Arch. Math. 54 (1990) pages 372-375. Conway was able to include a version in his book The Sensual Quadratic Form. A less Conwayish description, including just coefficients, in in Quaternary Quadratic Forms by Gordon L. Nipp. There we go, page 1 and then again on page 110, in genus 4: the final two forms with minimum 2, after which there are three forms with minimum 3 and that is the end of that genus and that discriminiminiminant.

I think you probably can decide this without tears, but just because your matrices are positive definite. Notice that, if $x$ is a row vector of integers, $$ x A x^T = x Q B Q^T x^T = y B y^T, $$ where we take $y = x Q.$ So, your two matrices have the same determinants, but they also have the same $\mathbb Z$-minimum. Indeed, I think they have the same "successive minima," and I would not be surprised if that were sufficient as well. In any case, you ought to get Gruber and Lekkerkerker. Note that emaphasis is switched from the matrix $A$ to the compact set with nonempty interior, $$ x A x^T \leq 1. $$

Now, why is is possible to find the $\mathbb Z$-minimum, which is some peculiar real number in your setting? Begin with $ m $ being the smallest diagonal entry of $A.$ The set of integer vectors $x$ with $ x A x^T \leq m$ is bounded by eigenvalue considerations (I actually do a little thing with Lagrange multipliers to get a separate bound for each coordinate $x_i.$) This gives a cube or rectangular box depending how you do it, leading to a finite set of integer vectors. So finding the $\mathbb Z$-minimum is a finite check.

I have seen this problem in infinite detail only for indefinite binary forms, see Cusick and Flahive

Alright, I am remembering things. Minkowski reduction is not definitive even for integral quadratic forms for dimensions at least three. It is quite good in dimensions 3,4. Alexander Schiemman made an improvement to Eisenstein's reduction in dimension 3, and wrote it up allowing for real entries, Mathematische Annalen, volume 308 (1997) pages 507-517, Ternary Positive Definite Quadratic Forms are determined by their Theta Series.

I think you probably can decide this without tears, but just because your matrices are positive definite. Notice that, if $x$ is a row vector of integers, $$ x A x^T = x Q B Q^T x^T = y B y^T, $$ where we take $y = x Q.$ So, your two matrices have the same determinants, but they also have the same $\mathbb Z$-minimum. Indeed, I think they have the same "successive minima," and I would not be surprised if that were sufficient as well. In any case, you ought to get Gruber and Lekkerkerker. Note that emaphasis is switched from the matrix $A$ to the compact set with nonempty interior, $$ x A x^T \leq 1. $$

Now, why is is possible to find the $\mathbb Z$-minimum, which is some peculiar real number in your setting? Begin with $ m $ being the smallest diagonal entry of $A.$ The set of integer vectors $x$ with $ x A x^T \leq m$ is bounded by eigenvalue considerations (I actually do a little thing with Lagrange multipliers to get a separate bound for each coordinate $x_i.$) This gives a cube or rectangular box depending how you do it, leading to a finite set of integer vectors. So finding the $\mathbb Z$-minimum is a finite check.

I have seen this problem in infinite detail only for indefinite binary forms, see Cusick and Flahive

Alright, I am remembering things. Minkowski reduction is not definitive even for integral quadratic forms for dimensions at least three. It is quite good in dimensions 3,4. Alexander Schiemman made an improvement to Eisenstein's reduction in dimension 3, and wrote it up allowing for real entries, Mathematische Annalen, volume 308 (1997) pages 507-517, Ternary Positive Definite Quadratic Forms are determined by their Theta Series.

EIT, 3:52 pm Pacific: Some of Schiemann's favorite themes give the simplest view of this. By definiteness, you can make two lists of represented values up to some bound, where "represented" means by an integral vector $x$ as above. Although these are real numbers, let me still call each list a "theta series," because we are able to count the number of times each real number is represented. Now, it is necessary that the two matrices have the same "theta series" in order to be $\mathbb Z$-equivalent. However, if the dimension is 4 or larger, this is not sufficient. The original examples of disagreement (same theta series but not equivalent) were by Milnor, dimension 16. The general topic is loosely under the heading of "Can you hear the shape of a drum?" Anyway, there was progress lowering the dimension, especially by Kneser and then Kitaoka. Finally, Schiemann found a four-dimensional pair, Arch. Math. 54 (1990) pages 372-375. Conway was able to include a version in his book The Sensual Quadratic Form. A less Conwayish description, including just coefficients, in in Quaternary Quadratic Forms by Gordon L. Nipp.