How can I give an explicit presentation of the ring of differential operators over a smooth variety?
$begingroup$
I'm trying to find a presentation of the ring of differential operators for a smooth affine and smooth projective variety. For example, consider the varieties
begin{align*}
X = textbf{Spec}left( R = frac{mathbb{C}[x,y]}{x^2 + y^2 - 1}right) && Y = textbf{Proj}left( S = frac{mathbb{C}[x,y,z]}{x^4 + y^4 + z^4}right)
end{align*}
In the first case, I can find the module of vector fields by computing the kahler differentials, and then computing the dual
$$
T_X = text{Hom}_Rleft(Omega_{R/mathbb{C}},Rright) = text{Hom}_Rleft(frac{Rdxoplus Rdy}{xdx + ydy},Rright) = frac{Rpartial_xoplus Rpartial_y}{xpartial_x + ypartial_y}
$$
and in the second case I can dualize the conormal sequence on Macaulay2.
How can I find a presentation for $D_X$ and $D_Y$? Originally I guessed that $D_X$ could be presented as
$$
frac{R[partial_x,partial_y]}{(xpartial_x + ypartial_y)}
$$
but the operator on the bottom does not preserve the ideal.
algebraic-geometry modules differential-operators d-modules
asked Aug 18 '16 at 22:22
54321user54321user
1,273622
$endgroup$
add a comment |
$begingroup$
I'm trying to find a presentation of the ring of differential operators for a smooth affine and smooth projective variety. For example, consider the varieties
begin{align*}
X = textbf{Spec}left( R = frac{mathbb{C}[x,y]}{x^2 + y^2 - 1}right) && Y = textbf{Proj}left( S = frac{mathbb{C}[x,y,z]}{x^4 + y^4 + z^4}right)
end{align*}
In the first case, I can find the module of vector fields by computing the kahler differentials, and then computing the dual
$$
T_X = text{Hom}_Rleft(Omega_{R/mathbb{C}},Rright) = text{Hom}_Rleft(frac{Rdxoplus Rdy}{xdx + ydy},Rright) = frac{Rpartial_xoplus Rpartial_y}{xpartial_x + ypartial_y}
$$
and in the second case I can dualize the conormal sequence on Macaulay2.
How can I find a presentation for $D_X$ and $D_Y$? Originally I guessed that $D_X$ could be presented as
$$
frac{R[partial_x,partial_y]}{(xpartial_x + ypartial_y)}
$$
but the operator on the bottom does not preserve the ideal.
algebraic-geometry modules differential-operators d-modules
asked Aug 18 '16 at 22:22
54321user54321user
1,273622
$endgroup$
add a comment |
$begingroup$
I'm trying to find a presentation of the ring of differential operators for a smooth affine and smooth projective variety. For example, consider the varieties
begin{align*}
X = textbf{Spec}left( R = frac{mathbb{C}[x,y]}{x^2 + y^2 - 1}right) && Y = textbf{Proj}left( S = frac{mathbb{C}[x,y,z]}{x^4 + y^4 + z^4}right)
end{align*}
In the first case, I can find the module of vector fields by computing the kahler differentials, and then computing the dual
$$
T_X = text{Hom}_Rleft(Omega_{R/mathbb{C}},Rright) = text{Hom}_Rleft(frac{Rdxoplus Rdy}{xdx + ydy},Rright) = frac{Rpartial_xoplus Rpartial_y}{xpartial_x + ypartial_y}
$$
and in the second case I can dualize the conormal sequence on Macaulay2.
How can I find a presentation for $D_X$ and $D_Y$? Originally I guessed that $D_X$ could be presented as
$$
frac{R[partial_x,partial_y]}{(xpartial_x + ypartial_y)}
$$
but the operator on the bottom does not preserve the ideal.
algebraic-geometry modules differential-operators d-modules
asked Aug 18 '16 at 22:22
54321user54321user
1,273622
$endgroup$
I'm trying to find a presentation of the ring of differential operators for a smooth affine and smooth projective variety. For example, consider the varieties
begin{align*}
X = textbf{Spec}left( R = frac{mathbb{C}[x,y]}{x^2 + y^2 - 1}right) && Y = textbf{Proj}left( S = frac{mathbb{C}[x,y,z]}{x^4 + y^4 + z^4}right)
end{align*}
In the first case, I can find the module of vector fields by computing the kahler differentials, and then computing the dual
$$
T_X = text{Hom}_Rleft(Omega_{R/mathbb{C}},Rright) = text{Hom}_Rleft(frac{Rdxoplus Rdy}{xdx + ydy},Rright) = frac{Rpartial_xoplus Rpartial_y}{xpartial_x + ypartial_y}
$$
and in the second case I can dualize the conormal sequence on Macaulay2.
How can I find a presentation for $D_X$ and $D_Y$? Originally I guessed that $D_X$ could be presented as
$$
frac{R[partial_x,partial_y]}{(xpartial_x + ypartial_y)}
$$
but the operator on the bottom does not preserve the ideal.
algebraic-geometry modules differential-operators d-modules
algebraic-geometry modules differential-operators d-modules
asked Aug 18 '16 at 22:22
54321user54321user
1,273622
asked Aug 18 '16 at 22:22
54321user54321user
1,273622
asked Aug 18 '16 at 22:22
54321user54321user
1,273622
asked Aug 18 '16 at 22:22
54321user54321user
1,273622
asked Aug 18 '16 at 22:22
54321user54321user
1,273622
1,273622
add a comment |
add a comment |
1 Answer
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I think there are two mistakes made by the OP which I will correct below.
In the ring of differential operators $D$, the element $x$ actually refers to the operator of multiplication by $x$. This does not commute with $partial_x$, instead $partial_x(xt) = t + xpartial_xt$ or $[partial_x,x] = 1$. Hence you cannot do $R[partial_x,partial_y]$, which would be a commutative ring. For instance, $$D(Bbb C[x_i]) = Bbb Clangle x_i,partial_irangle/([x_i,x_j],[x_i,partial_j] - delta_{ij}rangle.$$
Now you need to figure out the extra relations corresponding to $x^2 + y^2 - 1$ on $X$.
As you mentioned, your line $$Hom(Rdxoplus Rdy/(2x,dx + 2y,dy), R) overset{!}{=} Rpartial_xoplus Rpartial_y/(2xpartial_x + 2ypartial_y)$$ is not correct and you cannot "just switch" from tangent vectors to cotangent vectors.
So how do you compute it? What you need is something such which evaluates to 0 on $2xdx + 2ydy$. So, the solution is $ypartial_x - xpartial_y$. So $D(X) = A/(x^2+y^2-1)$ where $A$ is the subring: $$A = mathbb Clangle x,y,ypartial_x - xpartial_yrangle subset D(mathbb C[x,y])$$
You can get a presentation by combining this with the presentation above (or see the comments below this answer).
Note that this method works because $X$ is smooth. When $X$ is not smooth, you can't simply look at the ring generated by $mathcal O_X$ and $text{Der}(mathcal O_X,mathcal O_X)$, as is discussed for instance here https://cornellmath.wordpress.com/2007/09/09/d-module-basics-ii/
answered Dec 14 '18 at 5:02
BenBen
4,198617
$endgroup$
$begingroup$
Wouldn't the tangent bundle $mathscr{T}_{X / mathbb{C}}$ be equivalent to the morphisms $mathscr{O}_X cdot dx oplus mathscr{O}_X cdot dy to mathscr{O}_X$ which annihilate $x,dx + y,dy$? So, something like $mathscr{O}_X cdot (-y partial_x + x partial_y)$.
$endgroup$
– Daniel Schepler
Dec 14 '18 at 16:05
$begingroup$
@DanielSchepler Yeah that's what I was thinking - did I do it wrong?
$endgroup$
– Ben
Dec 15 '18 at 10:30
$begingroup$
Oh, OK - I got confused because in the next line you went back to a quotient for the differential operator ring. I'd expect the sheaf of diff ops to have a presentation something like $mathscr{O}_X langle partial_{theta} mid [f, partial_{theta}] = -y frac{partial f}{partial x} + x frac{partial f}{partial y} mathrm{~for~}finmathscr{O}_X rangle$. So the global sections would be something like $mathbb{C} langle x, y, partial_{theta} mid x^2+y^2=1, [partial_{theta}, x] = -y, [partial_{theta}, y] = x rangle$.
$endgroup$
– Daniel Schepler
Dec 15 '18 at 16:19
$begingroup$
@DanielSchepler Oh I see I forgot to put $x^2 +y^2 = 1$ in. Thanks!
$endgroup$
– Ben
Dec 16 '18 at 4:05
$begingroup$
@DanielSchepler Ah now I see what you are saying. Thank you!
$endgroup$
– Ben
Dec 16 '18 at 4:12
add a comment |
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1 Answer
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$begingroup$
I think there are two mistakes made by the OP which I will correct below.
In the ring of differential operators $D$, the element $x$ actually refers to the operator of multiplication by $x$. This does not commute with $partial_x$, instead $partial_x(xt) = t + xpartial_xt$ or $[partial_x,x] = 1$. Hence you cannot do $R[partial_x,partial_y]$, which would be a commutative ring. For instance, $$D(Bbb C[x_i]) = Bbb Clangle x_i,partial_irangle/([x_i,x_j],[x_i,partial_j] - delta_{ij}rangle.$$
Now you need to figure out the extra relations corresponding to $x^2 + y^2 - 1$ on $X$.
As you mentioned, your line $$Hom(Rdxoplus Rdy/(2x,dx + 2y,dy), R) overset{!}{=} Rpartial_xoplus Rpartial_y/(2xpartial_x + 2ypartial_y)$$ is not correct and you cannot "just switch" from tangent vectors to cotangent vectors.
So how do you compute it? What you need is something such which evaluates to 0 on $2xdx + 2ydy$. So, the solution is $ypartial_x - xpartial_y$. So $D(X) = A/(x^2+y^2-1)$ where $A$ is the subring: $$A = mathbb Clangle x,y,ypartial_x - xpartial_yrangle subset D(mathbb C[x,y])$$
You can get a presentation by combining this with the presentation above (or see the comments below this answer).
Note that this method works because $X$ is smooth. When $X$ is not smooth, you can't simply look at the ring generated by $mathcal O_X$ and $text{Der}(mathcal O_X,mathcal O_X)$, as is discussed for instance here https://cornellmath.wordpress.com/2007/09/09/d-module-basics-ii/
answered Dec 14 '18 at 5:02
BenBen
4,198617
$endgroup$
$begingroup$
Wouldn't the tangent bundle $mathscr{T}_{X / mathbb{C}}$ be equivalent to the morphisms $mathscr{O}_X cdot dx oplus mathscr{O}_X cdot dy to mathscr{O}_X$ which annihilate $x,dx + y,dy$? So, something like $mathscr{O}_X cdot (-y partial_x + x partial_y)$.
$endgroup$
– Daniel Schepler
Dec 14 '18 at 16:05
$begingroup$
@DanielSchepler Yeah that's what I was thinking - did I do it wrong?
$endgroup$
– Ben
Dec 15 '18 at 10:30
$begingroup$
Oh, OK - I got confused because in the next line you went back to a quotient for the differential operator ring. I'd expect the sheaf of diff ops to have a presentation something like $mathscr{O}_X langle partial_{theta} mid [f, partial_{theta}] = -y frac{partial f}{partial x} + x frac{partial f}{partial y} mathrm{~for~}finmathscr{O}_X rangle$. So the global sections would be something like $mathbb{C} langle x, y, partial_{theta} mid x^2+y^2=1, [partial_{theta}, x] = -y, [partial_{theta}, y] = x rangle$.
$endgroup$
– Daniel Schepler
Dec 15 '18 at 16:19
$begingroup$
@DanielSchepler Oh I see I forgot to put $x^2 +y^2 = 1$ in. Thanks!
$endgroup$
– Ben
Dec 16 '18 at 4:05
$begingroup$
@DanielSchepler Ah now I see what you are saying. Thank you!
$endgroup$
– Ben
Dec 16 '18 at 4:12
add a comment |
$begingroup$
I think there are two mistakes made by the OP which I will correct below.
In the ring of differential operators $D$, the element $x$ actually refers to the operator of multiplication by $x$. This does not commute with $partial_x$, instead $partial_x(xt) = t + xpartial_xt$ or $[partial_x,x] = 1$. Hence you cannot do $R[partial_x,partial_y]$, which would be a commutative ring. For instance, $$D(Bbb C[x_i]) = Bbb Clangle x_i,partial_irangle/([x_i,x_j],[x_i,partial_j] - delta_{ij}rangle.$$
Now you need to figure out the extra relations corresponding to $x^2 + y^2 - 1$ on $X$.
As you mentioned, your line $$Hom(Rdxoplus Rdy/(2x,dx + 2y,dy), R) overset{!}{=} Rpartial_xoplus Rpartial_y/(2xpartial_x + 2ypartial_y)$$ is not correct and you cannot "just switch" from tangent vectors to cotangent vectors.
So how do you compute it? What you need is something such which evaluates to 0 on $2xdx + 2ydy$. So, the solution is $ypartial_x - xpartial_y$. So $D(X) = A/(x^2+y^2-1)$ where $A$ is the subring: $$A = mathbb Clangle x,y,ypartial_x - xpartial_yrangle subset D(mathbb C[x,y])$$
You can get a presentation by combining this with the presentation above (or see the comments below this answer).
Note that this method works because $X$ is smooth. When $X$ is not smooth, you can't simply look at the ring generated by $mathcal O_X$ and $text{Der}(mathcal O_X,mathcal O_X)$, as is discussed for instance here https://cornellmath.wordpress.com/2007/09/09/d-module-basics-ii/
answered Dec 14 '18 at 5:02
BenBen
4,198617
$endgroup$
$begingroup$
Wouldn't the tangent bundle $mathscr{T}_{X / mathbb{C}}$ be equivalent to the morphisms $mathscr{O}_X cdot dx oplus mathscr{O}_X cdot dy to mathscr{O}_X$ which annihilate $x,dx + y,dy$? So, something like $mathscr{O}_X cdot (-y partial_x + x partial_y)$.
$endgroup$
– Daniel Schepler
Dec 14 '18 at 16:05
$begingroup$
@DanielSchepler Yeah that's what I was thinking - did I do it wrong?
$endgroup$
– Ben
Dec 15 '18 at 10:30
$begingroup$
Oh, OK - I got confused because in the next line you went back to a quotient for the differential operator ring. I'd expect the sheaf of diff ops to have a presentation something like $mathscr{O}_X langle partial_{theta} mid [f, partial_{theta}] = -y frac{partial f}{partial x} + x frac{partial f}{partial y} mathrm{~for~}finmathscr{O}_X rangle$. So the global sections would be something like $mathbb{C} langle x, y, partial_{theta} mid x^2+y^2=1, [partial_{theta}, x] = -y, [partial_{theta}, y] = x rangle$.
$endgroup$
– Daniel Schepler
Dec 15 '18 at 16:19
$begingroup$
@DanielSchepler Oh I see I forgot to put $x^2 +y^2 = 1$ in. Thanks!
$endgroup$
– Ben
Dec 16 '18 at 4:05
$begingroup$
@DanielSchepler Ah now I see what you are saying. Thank you!
$endgroup$
– Ben
Dec 16 '18 at 4:12
add a comment |
$begingroup$
I think there are two mistakes made by the OP which I will correct below.
In the ring of differential operators $D$, the element $x$ actually refers to the operator of multiplication by $x$. This does not commute with $partial_x$, instead $partial_x(xt) = t + xpartial_xt$ or $[partial_x,x] = 1$. Hence you cannot do $R[partial_x,partial_y]$, which would be a commutative ring. For instance, $$D(Bbb C[x_i]) = Bbb Clangle x_i,partial_irangle/([x_i,x_j],[x_i,partial_j] - delta_{ij}rangle.$$
Now you need to figure out the extra relations corresponding to $x^2 + y^2 - 1$ on $X$.
As you mentioned, your line $$Hom(Rdxoplus Rdy/(2x,dx + 2y,dy), R) overset{!}{=} Rpartial_xoplus Rpartial_y/(2xpartial_x + 2ypartial_y)$$ is not correct and you cannot "just switch" from tangent vectors to cotangent vectors.
So how do you compute it? What you need is something such which evaluates to 0 on $2xdx + 2ydy$. So, the solution is $ypartial_x - xpartial_y$. So $D(X) = A/(x^2+y^2-1)$ where $A$ is the subring: $$A = mathbb Clangle x,y,ypartial_x - xpartial_yrangle subset D(mathbb C[x,y])$$
You can get a presentation by combining this with the presentation above (or see the comments below this answer).
Note that this method works because $X$ is smooth. When $X$ is not smooth, you can't simply look at the ring generated by $mathcal O_X$ and $text{Der}(mathcal O_X,mathcal O_X)$, as is discussed for instance here https://cornellmath.wordpress.com/2007/09/09/d-module-basics-ii/
answered Dec 14 '18 at 5:02
BenBen
4,198617
$endgroup$
I think there are two mistakes made by the OP which I will correct below.
In the ring of differential operators $D$, the element $x$ actually refers to the operator of multiplication by $x$. This does not commute with $partial_x$, instead $partial_x(xt) = t + xpartial_xt$ or $[partial_x,x] = 1$. Hence you cannot do $R[partial_x,partial_y]$, which would be a commutative ring. For instance, $$D(Bbb C[x_i]) = Bbb Clangle x_i,partial_irangle/([x_i,x_j],[x_i,partial_j] - delta_{ij}rangle.$$
Now you need to figure out the extra relations corresponding to $x^2 + y^2 - 1$ on $X$.
As you mentioned, your line $$Hom(Rdxoplus Rdy/(2x,dx + 2y,dy), R) overset{!}{=} Rpartial_xoplus Rpartial_y/(2xpartial_x + 2ypartial_y)$$ is not correct and you cannot "just switch" from tangent vectors to cotangent vectors.
So how do you compute it? What you need is something such which evaluates to 0 on $2xdx + 2ydy$. So, the solution is $ypartial_x - xpartial_y$. So $D(X) = A/(x^2+y^2-1)$ where $A$ is the subring: $$A = mathbb Clangle x,y,ypartial_x - xpartial_yrangle subset D(mathbb C[x,y])$$
You can get a presentation by combining this with the presentation above (or see the comments below this answer).
Note that this method works because $X$ is smooth. When $X$ is not smooth, you can't simply look at the ring generated by $mathcal O_X$ and $text{Der}(mathcal O_X,mathcal O_X)$, as is discussed for instance here https://cornellmath.wordpress.com/2007/09/09/d-module-basics-ii/
answered Dec 14 '18 at 5:02
BenBen
4,198617
edited Dec 16 '18 at 9:15
answered Dec 14 '18 at 5:02
BenBen
4,198617
answered Dec 14 '18 at 5:02
BenBen
4,198617
answered Dec 14 '18 at 5:02
BenBen
4,198617
4,198617
$begingroup$
Wouldn't the tangent bundle $mathscr{T}_{X / mathbb{C}}$ be equivalent to the morphisms $mathscr{O}_X cdot dx oplus mathscr{O}_X cdot dy to mathscr{O}_X$ which annihilate $x,dx + y,dy$? So, something like $mathscr{O}_X cdot (-y partial_x + x partial_y)$.
$endgroup$
– Daniel Schepler
Dec 14 '18 at 16:05
$begingroup$
@DanielSchepler Yeah that's what I was thinking - did I do it wrong?
$endgroup$
– Ben
Dec 15 '18 at 10:30
$begingroup$
Oh, OK - I got confused because in the next line you went back to a quotient for the differential operator ring. I'd expect the sheaf of diff ops to have a presentation something like $mathscr{O}_X langle partial_{theta} mid [f, partial_{theta}] = -y frac{partial f}{partial x} + x frac{partial f}{partial y} mathrm{~for~}finmathscr{O}_X rangle$. So the global sections would be something like $mathbb{C} langle x, y, partial_{theta} mid x^2+y^2=1, [partial_{theta}, x] = -y, [partial_{theta}, y] = x rangle$.
$endgroup$
– Daniel Schepler
Dec 15 '18 at 16:19
$begingroup$
@DanielSchepler Oh I see I forgot to put $x^2 +y^2 = 1$ in. Thanks!
$endgroup$
– Ben
Dec 16 '18 at 4:05
$begingroup$
@DanielSchepler Ah now I see what you are saying. Thank you!
$endgroup$
– Ben
Dec 16 '18 at 4:12
add a comment |
$begingroup$
Wouldn't the tangent bundle $mathscr{T}_{X / mathbb{C}}$ be equivalent to the morphisms $mathscr{O}_X cdot dx oplus mathscr{O}_X cdot dy to mathscr{O}_X$ which annihilate $x,dx + y,dy$? So, something like $mathscr{O}_X cdot (-y partial_x + x partial_y)$.
$endgroup$
– Daniel Schepler
Dec 14 '18 at 16:05
$begingroup$
@DanielSchepler Yeah that's what I was thinking - did I do it wrong?
$endgroup$
– Ben
Dec 15 '18 at 10:30
$begingroup$
Oh, OK - I got confused because in the next line you went back to a quotient for the differential operator ring. I'd expect the sheaf of diff ops to have a presentation something like $mathscr{O}_X langle partial_{theta} mid [f, partial_{theta}] = -y frac{partial f}{partial x} + x frac{partial f}{partial y} mathrm{~for~}finmathscr{O}_X rangle$. So the global sections would be something like $mathbb{C} langle x, y, partial_{theta} mid x^2+y^2=1, [partial_{theta}, x] = -y, [partial_{theta}, y] = x rangle$.
$endgroup$
– Daniel Schepler
Dec 15 '18 at 16:19
$begingroup$
@DanielSchepler Oh I see I forgot to put $x^2 +y^2 = 1$ in. Thanks!
$endgroup$
– Ben
Dec 16 '18 at 4:05
$begingroup$
@DanielSchepler Ah now I see what you are saying. Thank you!
$endgroup$
– Ben
Dec 16 '18 at 4:12
$begingroup$
Wouldn't the tangent bundle $mathscr{T}_{X / mathbb{C}}$ be equivalent to the morphisms $mathscr{O}_X cdot dx oplus mathscr{O}_X cdot dy to mathscr{O}_X$ which annihilate $x,dx + y,dy$? So, something like $mathscr{O}_X cdot (-y partial_x + x partial_y)$.
$endgroup$
– Daniel Schepler
Dec 14 '18 at 16:05
$begingroup$
Wouldn't the tangent bundle $mathscr{T}_{X / mathbb{C}}$ be equivalent to the morphisms $mathscr{O}_X cdot dx oplus mathscr{O}_X cdot dy to mathscr{O}_X$ which annihilate $x,dx + y,dy$? So, something like $mathscr{O}_X cdot (-y partial_x + x partial_y)$.
$endgroup$
– Daniel Schepler
Dec 14 '18 at 16:05
$begingroup$
@DanielSchepler Yeah that's what I was thinking - did I do it wrong?
$endgroup$
– Ben
Dec 15 '18 at 10:30
$begingroup$
@DanielSchepler Yeah that's what I was thinking - did I do it wrong?
$endgroup$
– Ben
Dec 15 '18 at 10:30
$begingroup$
Oh, OK - I got confused because in the next line you went back to a quotient for the differential operator ring. I'd expect the sheaf of diff ops to have a presentation something like $mathscr{O}_X langle partial_{theta} mid [f, partial_{theta}] = -y frac{partial f}{partial x} + x frac{partial f}{partial y} mathrm{~for~}finmathscr{O}_X rangle$. So the global sections would be something like $mathbb{C} langle x, y, partial_{theta} mid x^2+y^2=1, [partial_{theta}, x] = -y, [partial_{theta}, y] = x rangle$.
$endgroup$
– Daniel Schepler
Dec 15 '18 at 16:19
$begingroup$
Oh, OK - I got confused because in the next line you went back to a quotient for the differential operator ring. I'd expect the sheaf of diff ops to have a presentation something like $mathscr{O}_X langle partial_{theta} mid [f, partial_{theta}] = -y frac{partial f}{partial x} + x frac{partial f}{partial y} mathrm{~for~}finmathscr{O}_X rangle$. So the global sections would be something like $mathbb{C} langle x, y, partial_{theta} mid x^2+y^2=1, [partial_{theta}, x] = -y, [partial_{theta}, y] = x rangle$.
$endgroup$
– Daniel Schepler
Dec 15 '18 at 16:19
$begingroup$
@DanielSchepler Oh I see I forgot to put $x^2 +y^2 = 1$ in. Thanks!
$endgroup$
– Ben
Dec 16 '18 at 4:05
$begingroup$
@DanielSchepler Oh I see I forgot to put $x^2 +y^2 = 1$ in. Thanks!
$endgroup$
– Ben
Dec 16 '18 at 4:05
$begingroup$
@DanielSchepler Ah now I see what you are saying. Thank you!
$endgroup$
– Ben
Dec 16 '18 at 4:12
$begingroup$
@DanielSchepler Ah now I see what you are saying. Thank you!
$endgroup$
– Ben
Dec 16 '18 at 4:12
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