Tangent bundle of sphere as a complex manifold












9












$begingroup$


I'm trying to show that the tangent bundle, $TS^n$ of the n-sphere $S^n$ is diffeomorphic to the set $sum z_i^2 = 1$ in $mathbb{C}^{n+1}$.



It's relatively straightforward to see that the tangent bundle of the sphere can be identified with:



$$TS^n = { (x_0,...,x_n,y_0,...,y_n) : x_i,y_i in mathbb{R}, sum x_i^2 = 1, sum x_i y_i = 0 }$$



Now to show this diffeomorphism I tried the natural thing of writing $z_j = x_j + iy_j$ but now we have $sum z_j^2 = 1 - sum y_i^2$ so it only lies in the required subspace if we restrict the tangent spaces of the sphere. I'm wondering how to write down a different map that does this?



I'm also a little concerned about how to show such a map is a diffeomorphism, how could I show that the identification I've made above as the tangent bundle embedded in $mathbb{R}^{2(n+1)}$ is smooth? It's probably obvious but I'm struggling to see it!



Thanks for any help






differential-geometry manifolds differential-topology smooth-manifolds







share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe I'm wrong, but shouldn't it be $sum x_i^2=1$?
    $endgroup$
    – B. Pasternak
    May 14 '16 at 12:22










  • $begingroup$
    Wait, why are $x_i$ and $y_i$ in $mathbb{R}^{n+1}$?
    $endgroup$
    – B. Pasternak
    May 14 '16 at 12:25










  • $begingroup$
    I think it definitively should be $x_i,y_iinmathbb{R}$ and $sum x_i^2=1$.
    $endgroup$
    – B. Pasternak
    May 14 '16 at 12:30










  • $begingroup$
    Thanks for your comments, yes there were a couple of typos which I've now corrected!
    $endgroup$
    – Wooster
    May 14 '16 at 12:58






  • 1




    $begingroup$
    Yes, you mean the hyperquadric $sum z_j^2 = 1$ in $Bbb C^{n+1}$. This is very non-compact and hardly a unit circle or unit sphere.
    $endgroup$
    – Ted Shifrin
    May 14 '16 at 17:00


















9












$begingroup$


I'm trying to show that the tangent bundle, $TS^n$ of the n-sphere $S^n$ is diffeomorphic to the set $sum z_i^2 = 1$ in $mathbb{C}^{n+1}$.



It's relatively straightforward to see that the tangent bundle of the sphere can be identified with:



$$TS^n = { (x_0,...,x_n,y_0,...,y_n) : x_i,y_i in mathbb{R}, sum x_i^2 = 1, sum x_i y_i = 0 }$$



Now to show this diffeomorphism I tried the natural thing of writing $z_j = x_j + iy_j$ but now we have $sum z_j^2 = 1 - sum y_i^2$ so it only lies in the required subspace if we restrict the tangent spaces of the sphere. I'm wondering how to write down a different map that does this?



I'm also a little concerned about how to show such a map is a diffeomorphism, how could I show that the identification I've made above as the tangent bundle embedded in $mathbb{R}^{2(n+1)}$ is smooth? It's probably obvious but I'm struggling to see it!



Thanks for any help






differential-geometry manifolds differential-topology smooth-manifolds







share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe I'm wrong, but shouldn't it be $sum x_i^2=1$?
    $endgroup$
    – B. Pasternak
    May 14 '16 at 12:22










  • $begingroup$
    Wait, why are $x_i$ and $y_i$ in $mathbb{R}^{n+1}$?
    $endgroup$
    – B. Pasternak
    May 14 '16 at 12:25










  • $begingroup$
    I think it definitively should be $x_i,y_iinmathbb{R}$ and $sum x_i^2=1$.
    $endgroup$
    – B. Pasternak
    May 14 '16 at 12:30










  • $begingroup$
    Thanks for your comments, yes there were a couple of typos which I've now corrected!
    $endgroup$
    – Wooster
    May 14 '16 at 12:58






  • 1




    $begingroup$
    Yes, you mean the hyperquadric $sum z_j^2 = 1$ in $Bbb C^{n+1}$. This is very non-compact and hardly a unit circle or unit sphere.
    $endgroup$
    – Ted Shifrin
    May 14 '16 at 17:00
















9












9








9


9



$begingroup$


I'm trying to show that the tangent bundle, $TS^n$ of the n-sphere $S^n$ is diffeomorphic to the set $sum z_i^2 = 1$ in $mathbb{C}^{n+1}$.



It's relatively straightforward to see that the tangent bundle of the sphere can be identified with:



$$TS^n = { (x_0,...,x_n,y_0,...,y_n) : x_i,y_i in mathbb{R}, sum x_i^2 = 1, sum x_i y_i = 0 }$$



Now to show this diffeomorphism I tried the natural thing of writing $z_j = x_j + iy_j$ but now we have $sum z_j^2 = 1 - sum y_i^2$ so it only lies in the required subspace if we restrict the tangent spaces of the sphere. I'm wondering how to write down a different map that does this?



I'm also a little concerned about how to show such a map is a diffeomorphism, how could I show that the identification I've made above as the tangent bundle embedded in $mathbb{R}^{2(n+1)}$ is smooth? It's probably obvious but I'm struggling to see it!



Thanks for any help






differential-geometry manifolds differential-topology smooth-manifolds







share|cite|improve this question











$endgroup$




I'm trying to show that the tangent bundle, $TS^n$ of the n-sphere $S^n$ is diffeomorphic to the set $sum z_i^2 = 1$ in $mathbb{C}^{n+1}$.



It's relatively straightforward to see that the tangent bundle of the sphere can be identified with:



$$TS^n = { (x_0,...,x_n,y_0,...,y_n) : x_i,y_i in mathbb{R}, sum x_i^2 = 1, sum x_i y_i = 0 }$$



Now to show this diffeomorphism I tried the natural thing of writing $z_j = x_j + iy_j$ but now we have $sum z_j^2 = 1 - sum y_i^2$ so it only lies in the required subspace if we restrict the tangent spaces of the sphere. I'm wondering how to write down a different map that does this?



I'm also a little concerned about how to show such a map is a diffeomorphism, how could I show that the identification I've made above as the tangent bundle embedded in $mathbb{R}^{2(n+1)}$ is smooth? It's probably obvious but I'm struggling to see it!



Thanks for any help






differential-geometry manifolds differential-topology smooth-manifolds




differential-geometry manifolds differential-topology smooth-manifolds






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 15 '16 at 8:05







Wooster

















asked May 14 '16 at 11:36









WoosterWooster

1,415936




1,415936












  • $begingroup$
    Maybe I'm wrong, but shouldn't it be $sum x_i^2=1$?
    $endgroup$
    – B. Pasternak
    May 14 '16 at 12:22










  • $begingroup$
    Wait, why are $x_i$ and $y_i$ in $mathbb{R}^{n+1}$?
    $endgroup$
    – B. Pasternak
    May 14 '16 at 12:25










  • $begingroup$
    I think it definitively should be $x_i,y_iinmathbb{R}$ and $sum x_i^2=1$.
    $endgroup$
    – B. Pasternak
    May 14 '16 at 12:30










  • $begingroup$
    Thanks for your comments, yes there were a couple of typos which I've now corrected!
    $endgroup$
    – Wooster
    May 14 '16 at 12:58






  • 1




    $begingroup$
    Yes, you mean the hyperquadric $sum z_j^2 = 1$ in $Bbb C^{n+1}$. This is very non-compact and hardly a unit circle or unit sphere.
    $endgroup$
    – Ted Shifrin
    May 14 '16 at 17:00




















  • $begingroup$
    Maybe I'm wrong, but shouldn't it be $sum x_i^2=1$?
    $endgroup$
    – B. Pasternak
    May 14 '16 at 12:22










  • $begingroup$
    Wait, why are $x_i$ and $y_i$ in $mathbb{R}^{n+1}$?
    $endgroup$
    – B. Pasternak
    May 14 '16 at 12:25










  • $begingroup$
    I think it definitively should be $x_i,y_iinmathbb{R}$ and $sum x_i^2=1$.
    $endgroup$
    – B. Pasternak
    May 14 '16 at 12:30










  • $begingroup$
    Thanks for your comments, yes there were a couple of typos which I've now corrected!
    $endgroup$
    – Wooster
    May 14 '16 at 12:58






  • 1




    $begingroup$
    Yes, you mean the hyperquadric $sum z_j^2 = 1$ in $Bbb C^{n+1}$. This is very non-compact and hardly a unit circle or unit sphere.
    $endgroup$
    – Ted Shifrin
    May 14 '16 at 17:00


















$begingroup$
Maybe I'm wrong, but shouldn't it be $sum x_i^2=1$?
$endgroup$
– B. Pasternak
May 14 '16 at 12:22




$begingroup$
Maybe I'm wrong, but shouldn't it be $sum x_i^2=1$?
$endgroup$
– B. Pasternak
May 14 '16 at 12:22












$begingroup$
Wait, why are $x_i$ and $y_i$ in $mathbb{R}^{n+1}$?
$endgroup$
– B. Pasternak
May 14 '16 at 12:25




$begingroup$
Wait, why are $x_i$ and $y_i$ in $mathbb{R}^{n+1}$?
$endgroup$
– B. Pasternak
May 14 '16 at 12:25












$begingroup$
I think it definitively should be $x_i,y_iinmathbb{R}$ and $sum x_i^2=1$.
$endgroup$
– B. Pasternak
May 14 '16 at 12:30




$begingroup$
I think it definitively should be $x_i,y_iinmathbb{R}$ and $sum x_i^2=1$.
$endgroup$
– B. Pasternak
May 14 '16 at 12:30












$begingroup$
Thanks for your comments, yes there were a couple of typos which I've now corrected!
$endgroup$
– Wooster
May 14 '16 at 12:58




$begingroup$
Thanks for your comments, yes there were a couple of typos which I've now corrected!
$endgroup$
– Wooster
May 14 '16 at 12:58




1




1




$begingroup$
Yes, you mean the hyperquadric $sum z_j^2 = 1$ in $Bbb C^{n+1}$. This is very non-compact and hardly a unit circle or unit sphere.
$endgroup$
– Ted Shifrin
May 14 '16 at 17:00






$begingroup$
Yes, you mean the hyperquadric $sum z_j^2 = 1$ in $Bbb C^{n+1}$. This is very non-compact and hardly a unit circle or unit sphere.
$endgroup$
– Ted Shifrin
May 14 '16 at 17:00












1 Answer
1






active

oldest

votes


















7












$begingroup$

Let $Qsubseteqmathbf C^{n+1}$ be the affine quadric defined by the equation $sum z_i^2=1$. The map
$$
fcolon TS^nrightarrow Q
$$
defined by
$$
z=f(x,y)=xsqrt{1+||y||^2}+ysqrt{-1}
$$
does the job, where $||y||^2=sum y_i^2$. Indeed,
one has $f(x,y)in Q$ since
$$
sum_{i=0}^n z_i^2=sum_{i=0}^n x_i^2(1+||y||^2)-y^2_i+2x_iy_isqrt{-1}sqrt{1+||y||^2}=\
1+||y||^2-||y||^2+2sqrt{-1}sqrt{1+||y||^2}sum x_iy_i=1,
$$
for $(x,y)in TS^n$.



The map $f$ is a diffeomorphism since its inverse is
$$
gcolon Qrightarrow TS^n
$$
defined by
$$
g(z)=left(frac{x}{sqrt{1+||y||^2}}, yright),
$$
where $z=x+ysqrt{-1}$.
One has $g(z)in TS^n$ since
$$
||x||^2-||y||^2=1
$$
and
$$
2sqrt{-1}sum x_iy_i=0
$$
for $zin Q$.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Excellent: +1. I have already seen that calculation, but I don't remember where. Would you happen to know in what context these calculations arise?
    $endgroup$
    – Georges Elencwajg
    May 14 '16 at 23:59






  • 2




    $begingroup$
    @Georges Elencwajg: I don't know about the specific calculation, but the general context is the one of Totaro's good complexifications (see math.ucla.edu/~totaro/papers/public_html/complex.pdf). A good complexification of a smooth manifiold $M$ is a smooth affine real algebraic variety $X$ such that $X(mathbf R)$ is diffeomorphic to $M$, and the inclusion $X(mathbf R)rightarrow X(mathbf C)$ is a homotopy equivalence. The manifold $S^n$ is particularly simple example of a manifold admitting a good complexification.
    $endgroup$
    – Johannes Huisman
    May 15 '16 at 5:55






  • 1




    $begingroup$
    Dear Johannes: thanks a lot for your explanations and reference. I find it quite interesting that this rather elementary question is related to quite advanced research by a brilliant mathematician like Totaro, who seems to have a knack for addressing difficult mathematical problems through rather unsophisticated considerations.
    $endgroup$
    – Georges Elencwajg
    May 15 '16 at 6:49








  • 1




    $begingroup$
    there is a slight error in the equation (I won't mess with your post). In the last terms it is $dots 2x_iy_isqrt{-(1+|y|^2)}$ and not $2x_iy_isqrt{-1}$. That does not affect the result of course (the inner product of $x$ and $y$ is still $0$). Thanks for the beautiful answer
    $endgroup$
    – Jose
    Jan 8 at 6:57










  • $begingroup$
    @Jose You're right. Thank you!
    $endgroup$
    – Johannes Huisman
    Jan 8 at 9:28












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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

Let $Qsubseteqmathbf C^{n+1}$ be the affine quadric defined by the equation $sum z_i^2=1$. The map
$$
fcolon TS^nrightarrow Q
$$
defined by
$$
z=f(x,y)=xsqrt{1+||y||^2}+ysqrt{-1}
$$
does the job, where $||y||^2=sum y_i^2$. Indeed,
one has $f(x,y)in Q$ since
$$
sum_{i=0}^n z_i^2=sum_{i=0}^n x_i^2(1+||y||^2)-y^2_i+2x_iy_isqrt{-1}sqrt{1+||y||^2}=\
1+||y||^2-||y||^2+2sqrt{-1}sqrt{1+||y||^2}sum x_iy_i=1,
$$
for $(x,y)in TS^n$.



The map $f$ is a diffeomorphism since its inverse is
$$
gcolon Qrightarrow TS^n
$$
defined by
$$
g(z)=left(frac{x}{sqrt{1+||y||^2}}, yright),
$$
where $z=x+ysqrt{-1}$.
One has $g(z)in TS^n$ since
$$
||x||^2-||y||^2=1
$$
and
$$
2sqrt{-1}sum x_iy_i=0
$$
for $zin Q$.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Excellent: +1. I have already seen that calculation, but I don't remember where. Would you happen to know in what context these calculations arise?
    $endgroup$
    – Georges Elencwajg
    May 14 '16 at 23:59






  • 2




    $begingroup$
    @Georges Elencwajg: I don't know about the specific calculation, but the general context is the one of Totaro's good complexifications (see math.ucla.edu/~totaro/papers/public_html/complex.pdf). A good complexification of a smooth manifiold $M$ is a smooth affine real algebraic variety $X$ such that $X(mathbf R)$ is diffeomorphic to $M$, and the inclusion $X(mathbf R)rightarrow X(mathbf C)$ is a homotopy equivalence. The manifold $S^n$ is particularly simple example of a manifold admitting a good complexification.
    $endgroup$
    – Johannes Huisman
    May 15 '16 at 5:55






  • 1




    $begingroup$
    Dear Johannes: thanks a lot for your explanations and reference. I find it quite interesting that this rather elementary question is related to quite advanced research by a brilliant mathematician like Totaro, who seems to have a knack for addressing difficult mathematical problems through rather unsophisticated considerations.
    $endgroup$
    – Georges Elencwajg
    May 15 '16 at 6:49








  • 1




    $begingroup$
    there is a slight error in the equation (I won't mess with your post). In the last terms it is $dots 2x_iy_isqrt{-(1+|y|^2)}$ and not $2x_iy_isqrt{-1}$. That does not affect the result of course (the inner product of $x$ and $y$ is still $0$). Thanks for the beautiful answer
    $endgroup$
    – Jose
    Jan 8 at 6:57










  • $begingroup$
    @Jose You're right. Thank you!
    $endgroup$
    – Johannes Huisman
    Jan 8 at 9:28
















7












$begingroup$

Let $Qsubseteqmathbf C^{n+1}$ be the affine quadric defined by the equation $sum z_i^2=1$. The map
$$
fcolon TS^nrightarrow Q
$$
defined by
$$
z=f(x,y)=xsqrt{1+||y||^2}+ysqrt{-1}
$$
does the job, where $||y||^2=sum y_i^2$. Indeed,
one has $f(x,y)in Q$ since
$$
sum_{i=0}^n z_i^2=sum_{i=0}^n x_i^2(1+||y||^2)-y^2_i+2x_iy_isqrt{-1}sqrt{1+||y||^2}=\
1+||y||^2-||y||^2+2sqrt{-1}sqrt{1+||y||^2}sum x_iy_i=1,
$$
for $(x,y)in TS^n$.



The map $f$ is a diffeomorphism since its inverse is
$$
gcolon Qrightarrow TS^n
$$
defined by
$$
g(z)=left(frac{x}{sqrt{1+||y||^2}}, yright),
$$
where $z=x+ysqrt{-1}$.
One has $g(z)in TS^n$ since
$$
||x||^2-||y||^2=1
$$
and
$$
2sqrt{-1}sum x_iy_i=0
$$
for $zin Q$.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Excellent: +1. I have already seen that calculation, but I don't remember where. Would you happen to know in what context these calculations arise?
    $endgroup$
    – Georges Elencwajg
    May 14 '16 at 23:59






  • 2




    $begingroup$
    @Georges Elencwajg: I don't know about the specific calculation, but the general context is the one of Totaro's good complexifications (see math.ucla.edu/~totaro/papers/public_html/complex.pdf). A good complexification of a smooth manifiold $M$ is a smooth affine real algebraic variety $X$ such that $X(mathbf R)$ is diffeomorphic to $M$, and the inclusion $X(mathbf R)rightarrow X(mathbf C)$ is a homotopy equivalence. The manifold $S^n$ is particularly simple example of a manifold admitting a good complexification.
    $endgroup$
    – Johannes Huisman
    May 15 '16 at 5:55






  • 1




    $begingroup$
    Dear Johannes: thanks a lot for your explanations and reference. I find it quite interesting that this rather elementary question is related to quite advanced research by a brilliant mathematician like Totaro, who seems to have a knack for addressing difficult mathematical problems through rather unsophisticated considerations.
    $endgroup$
    – Georges Elencwajg
    May 15 '16 at 6:49








  • 1




    $begingroup$
    there is a slight error in the equation (I won't mess with your post). In the last terms it is $dots 2x_iy_isqrt{-(1+|y|^2)}$ and not $2x_iy_isqrt{-1}$. That does not affect the result of course (the inner product of $x$ and $y$ is still $0$). Thanks for the beautiful answer
    $endgroup$
    – Jose
    Jan 8 at 6:57










  • $begingroup$
    @Jose You're right. Thank you!
    $endgroup$
    – Johannes Huisman
    Jan 8 at 9:28














7












7








7





$begingroup$

Let $Qsubseteqmathbf C^{n+1}$ be the affine quadric defined by the equation $sum z_i^2=1$. The map
$$
fcolon TS^nrightarrow Q
$$
defined by
$$
z=f(x,y)=xsqrt{1+||y||^2}+ysqrt{-1}
$$
does the job, where $||y||^2=sum y_i^2$. Indeed,
one has $f(x,y)in Q$ since
$$
sum_{i=0}^n z_i^2=sum_{i=0}^n x_i^2(1+||y||^2)-y^2_i+2x_iy_isqrt{-1}sqrt{1+||y||^2}=\
1+||y||^2-||y||^2+2sqrt{-1}sqrt{1+||y||^2}sum x_iy_i=1,
$$
for $(x,y)in TS^n$.



The map $f$ is a diffeomorphism since its inverse is
$$
gcolon Qrightarrow TS^n
$$
defined by
$$
g(z)=left(frac{x}{sqrt{1+||y||^2}}, yright),
$$
where $z=x+ysqrt{-1}$.
One has $g(z)in TS^n$ since
$$
||x||^2-||y||^2=1
$$
and
$$
2sqrt{-1}sum x_iy_i=0
$$
for $zin Q$.






share|cite|improve this answer











$endgroup$



Let $Qsubseteqmathbf C^{n+1}$ be the affine quadric defined by the equation $sum z_i^2=1$. The map
$$
fcolon TS^nrightarrow Q
$$
defined by
$$
z=f(x,y)=xsqrt{1+||y||^2}+ysqrt{-1}
$$
does the job, where $||y||^2=sum y_i^2$. Indeed,
one has $f(x,y)in Q$ since
$$
sum_{i=0}^n z_i^2=sum_{i=0}^n x_i^2(1+||y||^2)-y^2_i+2x_iy_isqrt{-1}sqrt{1+||y||^2}=\
1+||y||^2-||y||^2+2sqrt{-1}sqrt{1+||y||^2}sum x_iy_i=1,
$$
for $(x,y)in TS^n$.



The map $f$ is a diffeomorphism since its inverse is
$$
gcolon Qrightarrow TS^n
$$
defined by
$$
g(z)=left(frac{x}{sqrt{1+||y||^2}}, yright),
$$
where $z=x+ysqrt{-1}$.
One has $g(z)in TS^n$ since
$$
||x||^2-||y||^2=1
$$
and
$$
2sqrt{-1}sum x_iy_i=0
$$
for $zin Q$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 8 at 9:31

























answered May 14 '16 at 21:37









Johannes HuismanJohannes Huisman

3,146617




3,146617








  • 2




    $begingroup$
    Excellent: +1. I have already seen that calculation, but I don't remember where. Would you happen to know in what context these calculations arise?
    $endgroup$
    – Georges Elencwajg
    May 14 '16 at 23:59






  • 2




    $begingroup$
    @Georges Elencwajg: I don't know about the specific calculation, but the general context is the one of Totaro's good complexifications (see math.ucla.edu/~totaro/papers/public_html/complex.pdf). A good complexification of a smooth manifiold $M$ is a smooth affine real algebraic variety $X$ such that $X(mathbf R)$ is diffeomorphic to $M$, and the inclusion $X(mathbf R)rightarrow X(mathbf C)$ is a homotopy equivalence. The manifold $S^n$ is particularly simple example of a manifold admitting a good complexification.
    $endgroup$
    – Johannes Huisman
    May 15 '16 at 5:55






  • 1




    $begingroup$
    Dear Johannes: thanks a lot for your explanations and reference. I find it quite interesting that this rather elementary question is related to quite advanced research by a brilliant mathematician like Totaro, who seems to have a knack for addressing difficult mathematical problems through rather unsophisticated considerations.
    $endgroup$
    – Georges Elencwajg
    May 15 '16 at 6:49








  • 1




    $begingroup$
    there is a slight error in the equation (I won't mess with your post). In the last terms it is $dots 2x_iy_isqrt{-(1+|y|^2)}$ and not $2x_iy_isqrt{-1}$. That does not affect the result of course (the inner product of $x$ and $y$ is still $0$). Thanks for the beautiful answer
    $endgroup$
    – Jose
    Jan 8 at 6:57










  • $begingroup$
    @Jose You're right. Thank you!
    $endgroup$
    – Johannes Huisman
    Jan 8 at 9:28














  • 2




    $begingroup$
    Excellent: +1. I have already seen that calculation, but I don't remember where. Would you happen to know in what context these calculations arise?
    $endgroup$
    – Georges Elencwajg
    May 14 '16 at 23:59






  • 2




    $begingroup$
    @Georges Elencwajg: I don't know about the specific calculation, but the general context is the one of Totaro's good complexifications (see math.ucla.edu/~totaro/papers/public_html/complex.pdf). A good complexification of a smooth manifiold $M$ is a smooth affine real algebraic variety $X$ such that $X(mathbf R)$ is diffeomorphic to $M$, and the inclusion $X(mathbf R)rightarrow X(mathbf C)$ is a homotopy equivalence. The manifold $S^n$ is particularly simple example of a manifold admitting a good complexification.
    $endgroup$
    – Johannes Huisman
    May 15 '16 at 5:55






  • 1




    $begingroup$
    Dear Johannes: thanks a lot for your explanations and reference. I find it quite interesting that this rather elementary question is related to quite advanced research by a brilliant mathematician like Totaro, who seems to have a knack for addressing difficult mathematical problems through rather unsophisticated considerations.
    $endgroup$
    – Georges Elencwajg
    May 15 '16 at 6:49








  • 1




    $begingroup$
    there is a slight error in the equation (I won't mess with your post). In the last terms it is $dots 2x_iy_isqrt{-(1+|y|^2)}$ and not $2x_iy_isqrt{-1}$. That does not affect the result of course (the inner product of $x$ and $y$ is still $0$). Thanks for the beautiful answer
    $endgroup$
    – Jose
    Jan 8 at 6:57










  • $begingroup$
    @Jose You're right. Thank you!
    $endgroup$
    – Johannes Huisman
    Jan 8 at 9:28








2




2




$begingroup$
Excellent: +1. I have already seen that calculation, but I don't remember where. Would you happen to know in what context these calculations arise?
$endgroup$
– Georges Elencwajg
May 14 '16 at 23:59




$begingroup$
Excellent: +1. I have already seen that calculation, but I don't remember where. Would you happen to know in what context these calculations arise?
$endgroup$
– Georges Elencwajg
May 14 '16 at 23:59




2




2




$begingroup$
@Georges Elencwajg: I don't know about the specific calculation, but the general context is the one of Totaro's good complexifications (see math.ucla.edu/~totaro/papers/public_html/complex.pdf). A good complexification of a smooth manifiold $M$ is a smooth affine real algebraic variety $X$ such that $X(mathbf R)$ is diffeomorphic to $M$, and the inclusion $X(mathbf R)rightarrow X(mathbf C)$ is a homotopy equivalence. The manifold $S^n$ is particularly simple example of a manifold admitting a good complexification.
$endgroup$
– Johannes Huisman
May 15 '16 at 5:55




$begingroup$
@Georges Elencwajg: I don't know about the specific calculation, but the general context is the one of Totaro's good complexifications (see math.ucla.edu/~totaro/papers/public_html/complex.pdf). A good complexification of a smooth manifiold $M$ is a smooth affine real algebraic variety $X$ such that $X(mathbf R)$ is diffeomorphic to $M$, and the inclusion $X(mathbf R)rightarrow X(mathbf C)$ is a homotopy equivalence. The manifold $S^n$ is particularly simple example of a manifold admitting a good complexification.
$endgroup$
– Johannes Huisman
May 15 '16 at 5:55




1




1




$begingroup$
Dear Johannes: thanks a lot for your explanations and reference. I find it quite interesting that this rather elementary question is related to quite advanced research by a brilliant mathematician like Totaro, who seems to have a knack for addressing difficult mathematical problems through rather unsophisticated considerations.
$endgroup$
– Georges Elencwajg
May 15 '16 at 6:49






$begingroup$
Dear Johannes: thanks a lot for your explanations and reference. I find it quite interesting that this rather elementary question is related to quite advanced research by a brilliant mathematician like Totaro, who seems to have a knack for addressing difficult mathematical problems through rather unsophisticated considerations.
$endgroup$
– Georges Elencwajg
May 15 '16 at 6:49






1




1




$begingroup$
there is a slight error in the equation (I won't mess with your post). In the last terms it is $dots 2x_iy_isqrt{-(1+|y|^2)}$ and not $2x_iy_isqrt{-1}$. That does not affect the result of course (the inner product of $x$ and $y$ is still $0$). Thanks for the beautiful answer
$endgroup$
– Jose
Jan 8 at 6:57




$begingroup$
there is a slight error in the equation (I won't mess with your post). In the last terms it is $dots 2x_iy_isqrt{-(1+|y|^2)}$ and not $2x_iy_isqrt{-1}$. That does not affect the result of course (the inner product of $x$ and $y$ is still $0$). Thanks for the beautiful answer
$endgroup$
– Jose
Jan 8 at 6:57












$begingroup$
@Jose You're right. Thank you!
$endgroup$
– Johannes Huisman
Jan 8 at 9:28




$begingroup$
@Jose You're right. Thank you!
$endgroup$
– Johannes Huisman
Jan 8 at 9:28


















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