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Is there a characterization of automorphisms of degree-$2$ affine plane curves over $mathbbR$?
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Let $A$ denote a (nonsingular, irreducible) affine plane curve over $mathbbR$ given by the explicit equation
$$A : a_Ax^2+b_Axy+c_Ay^2+d_Ax+e_Ay+f_A$$
Question. Is there an explicit description of the automorphisms of $A$ in the category of real affine varieties?
algebraic-geometry
asked Jul 29 at 6:27
goblin
35.4k1153181
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up vote
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Let $A$ denote a (nonsingular, irreducible) affine plane curve over $mathbbR$ given by the explicit equation
$$A : a_Ax^2+b_Axy+c_Ay^2+d_Ax+e_Ay+f_A$$
Question. Is there an explicit description of the automorphisms of $A$ in the category of real affine varieties?
algebraic-geometry
asked Jul 29 at 6:27
goblin
35.4k1153181
1. Do you want it's automorphisms as an abstract variety or remembering the embedding in the plane? 2. How far short of your expectations does "compute discriminant, recognize what type of conic it is, remember automorphism group" fall?
– KReiser
Jul 29 at 6:47
@KReiser, isn't $(1)$ irrelevant? I thought the definition of a morphism $A rightarrow A$ was a pair of polynomials $P,Q in mathbbR[x,y]$ such that the induced map $(P,Q) : mathbbR^2 rightarrow mathbbR^2$ restricts to a function $A rightarrow A$. Am I wrong here?
– goblin
Jul 29 at 7:34
@KReiser, also $(2)$ sounds fine, but where can I find the automorphism groups computed explicitly? If I understand correctly, we just have to do this for $x^2-y^2=1$, $x^2+y^2 = 1$, and $y = x^2$ to get a complete solution, but I still don't know how to do it for those curves.
– goblin
Jul 29 at 7:35
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $A$ denote a (nonsingular, irreducible) affine plane curve over $mathbbR$ given by the explicit equation
$$A : a_Ax^2+b_Axy+c_Ay^2+d_Ax+e_Ay+f_A$$
Question. Is there an explicit description of the automorphisms of $A$ in the category of real affine varieties?
algebraic-geometry
asked Jul 29 at 6:27
goblin
35.4k1153181
Let $A$ denote a (nonsingular, irreducible) affine plane curve over $mathbbR$ given by the explicit equation
$$A : a_Ax^2+b_Axy+c_Ay^2+d_Ax+e_Ay+f_A$$
Question. Is there an explicit description of the automorphisms of $A$ in the category of real affine varieties?
algebraic-geometry
asked Jul 29 at 6:27
goblin
35.4k1153181
asked Jul 29 at 6:27
goblin
35.4k1153181
asked Jul 29 at 6:27
goblin
35.4k1153181
asked Jul 29 at 6:27
goblin
35.4k1153181
35.4k1153181
1. Do you want it's automorphisms as an abstract variety or remembering the embedding in the plane? 2. How far short of your expectations does "compute discriminant, recognize what type of conic it is, remember automorphism group" fall?
– KReiser
Jul 29 at 6:47
@KReiser, isn't $(1)$ irrelevant? I thought the definition of a morphism $A rightarrow A$ was a pair of polynomials $P,Q in mathbbR[x,y]$ such that the induced map $(P,Q) : mathbbR^2 rightarrow mathbbR^2$ restricts to a function $A rightarrow A$. Am I wrong here?
– goblin
Jul 29 at 7:34
@KReiser, also $(2)$ sounds fine, but where can I find the automorphism groups computed explicitly? If I understand correctly, we just have to do this for $x^2-y^2=1$, $x^2+y^2 = 1$, and $y = x^2$ to get a complete solution, but I still don't know how to do it for those curves.
– goblin
Jul 29 at 7:35
add a comment |Â
1. Do you want it's automorphisms as an abstract variety or remembering the embedding in the plane? 2. How far short of your expectations does "compute discriminant, recognize what type of conic it is, remember automorphism group" fall?
– KReiser
Jul 29 at 6:47
@KReiser, isn't $(1)$ irrelevant? I thought the definition of a morphism $A rightarrow A$ was a pair of polynomials $P,Q in mathbbR[x,y]$ such that the induced map $(P,Q) : mathbbR^2 rightarrow mathbbR^2$ restricts to a function $A rightarrow A$. Am I wrong here?
– goblin
Jul 29 at 7:34
@KReiser, also $(2)$ sounds fine, but where can I find the automorphism groups computed explicitly? If I understand correctly, we just have to do this for $x^2-y^2=1$, $x^2+y^2 = 1$, and $y = x^2$ to get a complete solution, but I still don't know how to do it for those curves.
– goblin
Jul 29 at 7:35
1. Do you want it's automorphisms as an abstract variety or remembering the embedding in the plane? 2. How far short of your expectations does "compute discriminant, recognize what type of conic it is, remember automorphism group" fall?
– KReiser
Jul 29 at 6:47
1. Do you want it's automorphisms as an abstract variety or remembering the embedding in the plane? 2. How far short of your expectations does "compute discriminant, recognize what type of conic it is, remember automorphism group" fall?
– KReiser
Jul 29 at 6:47
@KReiser, isn't $(1)$ irrelevant? I thought the definition of a morphism $A rightarrow A$ was a pair of polynomials $P,Q in mathbbR[x,y]$ such that the induced map $(P,Q) : mathbbR^2 rightarrow mathbbR^2$ restricts to a function $A rightarrow A$. Am I wrong here?
– goblin
Jul 29 at 7:34
@KReiser, isn't $(1)$ irrelevant? I thought the definition of a morphism $A rightarrow A$ was a pair of polynomials $P,Q in mathbbR[x,y]$ such that the induced map $(P,Q) : mathbbR^2 rightarrow mathbbR^2$ restricts to a function $A rightarrow A$. Am I wrong here?
– goblin
Jul 29 at 7:34
@KReiser, also $(2)$ sounds fine, but where can I find the automorphism groups computed explicitly? If I understand correctly, we just have to do this for $x^2-y^2=1$, $x^2+y^2 = 1$, and $y = x^2$ to get a complete solution, but I still don't know how to do it for those curves.
– goblin
Jul 29 at 7:35
@KReiser, also $(2)$ sounds fine, but where can I find the automorphism groups computed explicitly? If I understand correctly, we just have to do this for $x^2-y^2=1$, $x^2+y^2 = 1$, and $y = x^2$ to get a complete solution, but I still don't know how to do it for those curves.
– goblin
Jul 29 at 7:35
add a comment |Â
1 Answer
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We will compute the result for $Bbb C$, and note that an automorphism of $operatornameSpec Bbb R[x,y]/f$ is the same as an automorphism of $operatornameSpec Bbb C[x,y]/f$ with real coefficients. Over $Bbb C$, there are exactly two affine plane conics up to isomorphism: $xy=1, y=x^2$. So it suffices to compute the automorphism groups of each of these curves and do some work translating things back to the reals.
One preliminary result that I'll use without proof: $Aut(k(x))= PSL(2,k)$ acting by Mobius transformations.
Case 1: $y=x^2$. The coordinate algebra is isomorphic to $Bbb C[x]$ by writing $y=x^2$. Any automorphism of $k[x]$ extends to an automorphism of $k(x)$, and the automorphisms of $k(x)$ that preserve $k[x]$ are exactly the linear maps. So the automorphisms of $Bbb C[x]$ are exactly the linear maps $xmapsto ax+b$ with $aneq 0$. In order for these automorphisms to restrict the correct way to our curve, we just need $a,binBbb R$.
Case 2: $xy=1$. The coordinate algebra is isomorphic to $Bbb C[x,x^-1]$, and any automorphism is determined by it's action on $x$. Again, any automorphism of $k[x,x^-1]$ extends to an automorphism of $k(x)$, and the automorphisms of $k(x)$ that preserve $k[x,x^-1]$ are $xmapsto ax$ and $xmapsto cx^-1$ for $a,cneq 0$.
We need to do a little more work in case 2 to translate these automorphisms back to the land of the reals. In the case that our curve over $Bbb R$ is a hyperbola, we can make the change of variables $xmapsto (x-y),ymapsto (x+y)$ to translate between $xy=1$ and $x^2-y^2=1$. Recognizing this coordinate change as a rotation and performing some algebra, we see that the symmetries of $x^2-y^2=1$ are all of the form $beginpmatrix pm cosh(a) & sinh(a) \ pm sinh(a) & cosh(a) endpmatrix$ for $ain [0,2pi)$.
Similar logic with the change of variables $xmapsto x-iy, ymapsto x+iy$ shows that the automorphisms of $x^2+y^2=1$ are exactly the rotation matrices up to a reflection across any line through the origin, ie matrices of the form $beginpmatrix pmcos(a) & -sin(a) \ pmsin(a) & cos(a) endpmatrix$.
I wouldn't be surprised if there were a shorter proof, but this is the clearest approach I could think of while writing.
answered Jul 30 at 22:53
KReiser
7,45011230
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
We will compute the result for $Bbb C$, and note that an automorphism of $operatornameSpec Bbb R[x,y]/f$ is the same as an automorphism of $operatornameSpec Bbb C[x,y]/f$ with real coefficients. Over $Bbb C$, there are exactly two affine plane conics up to isomorphism: $xy=1, y=x^2$. So it suffices to compute the automorphism groups of each of these curves and do some work translating things back to the reals.
One preliminary result that I'll use without proof: $Aut(k(x))= PSL(2,k)$ acting by Mobius transformations.
Case 1: $y=x^2$. The coordinate algebra is isomorphic to $Bbb C[x]$ by writing $y=x^2$. Any automorphism of $k[x]$ extends to an automorphism of $k(x)$, and the automorphisms of $k(x)$ that preserve $k[x]$ are exactly the linear maps. So the automorphisms of $Bbb C[x]$ are exactly the linear maps $xmapsto ax+b$ with $aneq 0$. In order for these automorphisms to restrict the correct way to our curve, we just need $a,binBbb R$.
Case 2: $xy=1$. The coordinate algebra is isomorphic to $Bbb C[x,x^-1]$, and any automorphism is determined by it's action on $x$. Again, any automorphism of $k[x,x^-1]$ extends to an automorphism of $k(x)$, and the automorphisms of $k(x)$ that preserve $k[x,x^-1]$ are $xmapsto ax$ and $xmapsto cx^-1$ for $a,cneq 0$.
We need to do a little more work in case 2 to translate these automorphisms back to the land of the reals. In the case that our curve over $Bbb R$ is a hyperbola, we can make the change of variables $xmapsto (x-y),ymapsto (x+y)$ to translate between $xy=1$ and $x^2-y^2=1$. Recognizing this coordinate change as a rotation and performing some algebra, we see that the symmetries of $x^2-y^2=1$ are all of the form $beginpmatrix pm cosh(a) & sinh(a) \ pm sinh(a) & cosh(a) endpmatrix$ for $ain [0,2pi)$.
Similar logic with the change of variables $xmapsto x-iy, ymapsto x+iy$ shows that the automorphisms of $x^2+y^2=1$ are exactly the rotation matrices up to a reflection across any line through the origin, ie matrices of the form $beginpmatrix pmcos(a) & -sin(a) \ pmsin(a) & cos(a) endpmatrix$.
I wouldn't be surprised if there were a shorter proof, but this is the clearest approach I could think of while writing.
answered Jul 30 at 22:53
KReiser
7,45011230
add a comment |Â
up vote
1
down vote
We will compute the result for $Bbb C$, and note that an automorphism of $operatornameSpec Bbb R[x,y]/f$ is the same as an automorphism of $operatornameSpec Bbb C[x,y]/f$ with real coefficients. Over $Bbb C$, there are exactly two affine plane conics up to isomorphism: $xy=1, y=x^2$. So it suffices to compute the automorphism groups of each of these curves and do some work translating things back to the reals.
One preliminary result that I'll use without proof: $Aut(k(x))= PSL(2,k)$ acting by Mobius transformations.
Case 1: $y=x^2$. The coordinate algebra is isomorphic to $Bbb C[x]$ by writing $y=x^2$. Any automorphism of $k[x]$ extends to an automorphism of $k(x)$, and the automorphisms of $k(x)$ that preserve $k[x]$ are exactly the linear maps. So the automorphisms of $Bbb C[x]$ are exactly the linear maps $xmapsto ax+b$ with $aneq 0$. In order for these automorphisms to restrict the correct way to our curve, we just need $a,binBbb R$.
Case 2: $xy=1$. The coordinate algebra is isomorphic to $Bbb C[x,x^-1]$, and any automorphism is determined by it's action on $x$. Again, any automorphism of $k[x,x^-1]$ extends to an automorphism of $k(x)$, and the automorphisms of $k(x)$ that preserve $k[x,x^-1]$ are $xmapsto ax$ and $xmapsto cx^-1$ for $a,cneq 0$.
We need to do a little more work in case 2 to translate these automorphisms back to the land of the reals. In the case that our curve over $Bbb R$ is a hyperbola, we can make the change of variables $xmapsto (x-y),ymapsto (x+y)$ to translate between $xy=1$ and $x^2-y^2=1$. Recognizing this coordinate change as a rotation and performing some algebra, we see that the symmetries of $x^2-y^2=1$ are all of the form $beginpmatrix pm cosh(a) & sinh(a) \ pm sinh(a) & cosh(a) endpmatrix$ for $ain [0,2pi)$.
Similar logic with the change of variables $xmapsto x-iy, ymapsto x+iy$ shows that the automorphisms of $x^2+y^2=1$ are exactly the rotation matrices up to a reflection across any line through the origin, ie matrices of the form $beginpmatrix pmcos(a) & -sin(a) \ pmsin(a) & cos(a) endpmatrix$.
I wouldn't be surprised if there were a shorter proof, but this is the clearest approach I could think of while writing.
answered Jul 30 at 22:53
KReiser
7,45011230
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We will compute the result for $Bbb C$, and note that an automorphism of $operatornameSpec Bbb R[x,y]/f$ is the same as an automorphism of $operatornameSpec Bbb C[x,y]/f$ with real coefficients. Over $Bbb C$, there are exactly two affine plane conics up to isomorphism: $xy=1, y=x^2$. So it suffices to compute the automorphism groups of each of these curves and do some work translating things back to the reals.
One preliminary result that I'll use without proof: $Aut(k(x))= PSL(2,k)$ acting by Mobius transformations.
Case 1: $y=x^2$. The coordinate algebra is isomorphic to $Bbb C[x]$ by writing $y=x^2$. Any automorphism of $k[x]$ extends to an automorphism of $k(x)$, and the automorphisms of $k(x)$ that preserve $k[x]$ are exactly the linear maps. So the automorphisms of $Bbb C[x]$ are exactly the linear maps $xmapsto ax+b$ with $aneq 0$. In order for these automorphisms to restrict the correct way to our curve, we just need $a,binBbb R$.
Case 2: $xy=1$. The coordinate algebra is isomorphic to $Bbb C[x,x^-1]$, and any automorphism is determined by it's action on $x$. Again, any automorphism of $k[x,x^-1]$ extends to an automorphism of $k(x)$, and the automorphisms of $k(x)$ that preserve $k[x,x^-1]$ are $xmapsto ax$ and $xmapsto cx^-1$ for $a,cneq 0$.
We need to do a little more work in case 2 to translate these automorphisms back to the land of the reals. In the case that our curve over $Bbb R$ is a hyperbola, we can make the change of variables $xmapsto (x-y),ymapsto (x+y)$ to translate between $xy=1$ and $x^2-y^2=1$. Recognizing this coordinate change as a rotation and performing some algebra, we see that the symmetries of $x^2-y^2=1$ are all of the form $beginpmatrix pm cosh(a) & sinh(a) \ pm sinh(a) & cosh(a) endpmatrix$ for $ain [0,2pi)$.
Similar logic with the change of variables $xmapsto x-iy, ymapsto x+iy$ shows that the automorphisms of $x^2+y^2=1$ are exactly the rotation matrices up to a reflection across any line through the origin, ie matrices of the form $beginpmatrix pmcos(a) & -sin(a) \ pmsin(a) & cos(a) endpmatrix$.
I wouldn't be surprised if there were a shorter proof, but this is the clearest approach I could think of while writing.
answered Jul 30 at 22:53
KReiser
7,45011230
We will compute the result for $Bbb C$, and note that an automorphism of $operatornameSpec Bbb R[x,y]/f$ is the same as an automorphism of $operatornameSpec Bbb C[x,y]/f$ with real coefficients. Over $Bbb C$, there are exactly two affine plane conics up to isomorphism: $xy=1, y=x^2$. So it suffices to compute the automorphism groups of each of these curves and do some work translating things back to the reals.
One preliminary result that I'll use without proof: $Aut(k(x))= PSL(2,k)$ acting by Mobius transformations.
Case 1: $y=x^2$. The coordinate algebra is isomorphic to $Bbb C[x]$ by writing $y=x^2$. Any automorphism of $k[x]$ extends to an automorphism of $k(x)$, and the automorphisms of $k(x)$ that preserve $k[x]$ are exactly the linear maps. So the automorphisms of $Bbb C[x]$ are exactly the linear maps $xmapsto ax+b$ with $aneq 0$. In order for these automorphisms to restrict the correct way to our curve, we just need $a,binBbb R$.
Case 2: $xy=1$. The coordinate algebra is isomorphic to $Bbb C[x,x^-1]$, and any automorphism is determined by it's action on $x$. Again, any automorphism of $k[x,x^-1]$ extends to an automorphism of $k(x)$, and the automorphisms of $k(x)$ that preserve $k[x,x^-1]$ are $xmapsto ax$ and $xmapsto cx^-1$ for $a,cneq 0$.
We need to do a little more work in case 2 to translate these automorphisms back to the land of the reals. In the case that our curve over $Bbb R$ is a hyperbola, we can make the change of variables $xmapsto (x-y),ymapsto (x+y)$ to translate between $xy=1$ and $x^2-y^2=1$. Recognizing this coordinate change as a rotation and performing some algebra, we see that the symmetries of $x^2-y^2=1$ are all of the form $beginpmatrix pm cosh(a) & sinh(a) \ pm sinh(a) & cosh(a) endpmatrix$ for $ain [0,2pi)$.
Similar logic with the change of variables $xmapsto x-iy, ymapsto x+iy$ shows that the automorphisms of $x^2+y^2=1$ are exactly the rotation matrices up to a reflection across any line through the origin, ie matrices of the form $beginpmatrix pmcos(a) & -sin(a) \ pmsin(a) & cos(a) endpmatrix$.
I wouldn't be surprised if there were a shorter proof, but this is the clearest approach I could think of while writing.
answered Jul 30 at 22:53
KReiser
7,45011230
answered Jul 30 at 22:53
KReiser
7,45011230
answered Jul 30 at 22:53
KReiser
7,45011230
answered Jul 30 at 22:53
KReiser
7,45011230
7,45011230
add a comment |Â
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1. Do you want it's automorphisms as an abstract variety or remembering the embedding in the plane? 2. How far short of your expectations does "compute discriminant, recognize what type of conic it is, remember automorphism group" fall?
– KReiser
Jul 29 at 6:47
@KReiser, isn't $(1)$ irrelevant? I thought the definition of a morphism $A rightarrow A$ was a pair of polynomials $P,Q in mathbbR[x,y]$ such that the induced map $(P,Q) : mathbbR^2 rightarrow mathbbR^2$ restricts to a function $A rightarrow A$. Am I wrong here?
– goblin
Jul 29 at 7:34
@KReiser, also $(2)$ sounds fine, but where can I find the automorphism groups computed explicitly? If I understand correctly, we just have to do this for $x^2-y^2=1$, $x^2+y^2 = 1$, and $y = x^2$ to get a complete solution, but I still don't know how to do it for those curves.
– goblin
Jul 29 at 7:35