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Vít Tuček
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Let $f$ be a ternary quadratic form, say with real coefficients. Then there is an associated symmetric matrix to $f$, say $A_f$. The orthogonal group of $f$ is then the group

$$\displaystyle O_f = \{H \in \operatorname{GL}_3(\mathbb{R}): H^T A_f H = A\}.$$$$\displaystyle O_f = \{H \in \operatorname{GL}_3(\mathbb{R}): H^T A_f H = A_f\}.$$

Given two non-proportional ternary quadratic forms $f,g$, which we may assume are not singular, how do we determine the group $O_f \cap O_g$? Is there a way to canonically write down generators of this group?

Let $f$ be a ternary quadratic form, say with real coefficients. Then there is an associated symmetric matrix to $f$, say $A_f$. The orthogonal group of $f$ is then the group

$$\displaystyle O_f = \{H \in \operatorname{GL}_3(\mathbb{R}): H^T A_f H = A\}.$$

Given two non-proportional ternary quadratic forms $f,g$, which we may assume are not singular, how do we determine the group $O_f \cap O_g$? Is there a way to canonically write down generators of this group?

Let $f$ be a ternary quadratic form, say with real coefficients. Then there is an associated symmetric matrix to $f$, say $A_f$. The orthogonal group of $f$ is then the group

$$\displaystyle O_f = \{H \in \operatorname{GL}_3(\mathbb{R}): H^T A_f H = A_f\}.$$

Given two non-proportional ternary quadratic forms $f,g$, which we may assume are not singular, how do we determine the group $O_f \cap O_g$? Is there a way to canonically write down generators of this group?

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Stanley Yao Xiao
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Intersection of orthogonal groups

Let $f$ be a ternary quadratic form, say with real coefficients. Then there is an associated symmetric matrix to $f$, say $A_f$. The orthogonal group of $f$ is then the group

$$\displaystyle O_f = \{H \in \operatorname{GL}_3(\mathbb{R}): H^T A_f H = A\}.$$

Given two non-proportional ternary quadratic forms $f,g$, which we may assume are not singular, how do we determine the group $O_f \cap O_g$? Is there a way to canonically write down generators of this group?