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Is there an explanation of analogies between the cross-ratio and the Riemann curvature tensor?
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Define the cross-ratio of four real or complex numbers as follows:
$$[a,b,c,d] = frac(a-c)(b-d)(a-d)(b-c).$$
Then its logarithm has the same symmetries as the curvature tensor:
$$log[a,b,c,d] = -log[b,a,c,d] = -log[a,b,d,c] = log[c,d,a,b].$$
Moreover, if $[a,b,c,d] = lambda$, then $[b,c,a,d] = 1 - lambda^-1$ and $[c,a,b,d] = (1-lambda)^-1$, which implies an analog of the algebraic Bianchi identity:
$$log[a,b,c,d] + log[b,c,a,d] + log[c,a,b,d] = pi i.$$
Is there something behind these coincidences?
ag.algebraic-geometry riemannian-geometry
asked 5 hours ago
Ivan Izmestiev
3,3051237
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up vote
11
down vote
favorite
Define the cross-ratio of four real or complex numbers as follows:
$$[a,b,c,d] = frac(a-c)(b-d)(a-d)(b-c).$$
Then its logarithm has the same symmetries as the curvature tensor:
$$log[a,b,c,d] = -log[b,a,c,d] = -log[a,b,d,c] = log[c,d,a,b].$$
Moreover, if $[a,b,c,d] = lambda$, then $[b,c,a,d] = 1 - lambda^-1$ and $[c,a,b,d] = (1-lambda)^-1$, which implies an analog of the algebraic Bianchi identity:
$$log[a,b,c,d] + log[b,c,a,d] + log[c,a,b,d] = pi i.$$
Is there something behind these coincidences?
ag.algebraic-geometry riemannian-geometry
asked 5 hours ago
Ivan Izmestiev
3,3051237
add a comment |Â
up vote
11
down vote
favorite
up vote
11
down vote
favorite
Define the cross-ratio of four real or complex numbers as follows:
$$[a,b,c,d] = frac(a-c)(b-d)(a-d)(b-c).$$
Then its logarithm has the same symmetries as the curvature tensor:
$$log[a,b,c,d] = -log[b,a,c,d] = -log[a,b,d,c] = log[c,d,a,b].$$
Moreover, if $[a,b,c,d] = lambda$, then $[b,c,a,d] = 1 - lambda^-1$ and $[c,a,b,d] = (1-lambda)^-1$, which implies an analog of the algebraic Bianchi identity:
$$log[a,b,c,d] + log[b,c,a,d] + log[c,a,b,d] = pi i.$$
Is there something behind these coincidences?
ag.algebraic-geometry riemannian-geometry
asked 5 hours ago
Ivan Izmestiev
3,3051237
Define the cross-ratio of four real or complex numbers as follows:
$$[a,b,c,d] = frac(a-c)(b-d)(a-d)(b-c).$$
Then its logarithm has the same symmetries as the curvature tensor:
$$log[a,b,c,d] = -log[b,a,c,d] = -log[a,b,d,c] = log[c,d,a,b].$$
Moreover, if $[a,b,c,d] = lambda$, then $[b,c,a,d] = 1 - lambda^-1$ and $[c,a,b,d] = (1-lambda)^-1$, which implies an analog of the algebraic Bianchi identity:
$$log[a,b,c,d] + log[b,c,a,d] + log[c,a,b,d] = pi i.$$
Is there something behind these coincidences?
ag.algebraic-geometry riemannian-geometry
ag.algebraic-geometry riemannian-geometry
asked 5 hours ago
Ivan Izmestiev
3,3051237
asked 5 hours ago
Ivan Izmestiev
3,3051237
edited 4 hours ago
asked 5 hours ago
Ivan Izmestiev
3,3051237
asked 5 hours ago
Ivan Izmestiev
3,3051237
asked 5 hours ago
Ivan Izmestiev
3,3051237
3,3051237
add a comment |Â
add a comment |Â
1 Answer
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There is an indirect connection which goes via the representation theory of the symmetric group. The symmetries of the Riemann tensor are equivalent to saying that $R$ transforms according to the two-dimensional irreducible representation of $mathbb S_4$ corresponding to the partition $[2,2]$. On the other hand let us consider the moduli space $M_0,4$ parametrizing four distinct ordered points on the Riemann sphere up to Möbius transformations. The cohomology group $H^1(M_0,4,mathbf C)$ is $2$-dimensional, and transforms according to the representation $[2,2]$ under its natural action of $mathbb S_4$. But we may compute this cohomology group as the space of holomorphic $1$-forms on $M_0,4$ with at most logarithmic poles at infinity. Moreover, the cross-ratio $chi$ may be considered as a holomorphic function on $M_0,4$ and its logarithmic derivative $d log(chi)$ is such a 1-form with log poles. (Note that differentiating your analogue of the Bianchi identity gets rid of the $pi i$, so we really get something exactly like the usual Bianchi identity!)
answered 52 mins ago
Dan Petersen
24.4k268131
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
There is an indirect connection which goes via the representation theory of the symmetric group. The symmetries of the Riemann tensor are equivalent to saying that $R$ transforms according to the two-dimensional irreducible representation of $mathbb S_4$ corresponding to the partition $[2,2]$. On the other hand let us consider the moduli space $M_0,4$ parametrizing four distinct ordered points on the Riemann sphere up to Möbius transformations. The cohomology group $H^1(M_0,4,mathbf C)$ is $2$-dimensional, and transforms according to the representation $[2,2]$ under its natural action of $mathbb S_4$. But we may compute this cohomology group as the space of holomorphic $1$-forms on $M_0,4$ with at most logarithmic poles at infinity. Moreover, the cross-ratio $chi$ may be considered as a holomorphic function on $M_0,4$ and its logarithmic derivative $d log(chi)$ is such a 1-form with log poles. (Note that differentiating your analogue of the Bianchi identity gets rid of the $pi i$, so we really get something exactly like the usual Bianchi identity!)
answered 52 mins ago
Dan Petersen
24.4k268131
add a comment |Â
up vote
4
down vote
There is an indirect connection which goes via the representation theory of the symmetric group. The symmetries of the Riemann tensor are equivalent to saying that $R$ transforms according to the two-dimensional irreducible representation of $mathbb S_4$ corresponding to the partition $[2,2]$. On the other hand let us consider the moduli space $M_0,4$ parametrizing four distinct ordered points on the Riemann sphere up to Möbius transformations. The cohomology group $H^1(M_0,4,mathbf C)$ is $2$-dimensional, and transforms according to the representation $[2,2]$ under its natural action of $mathbb S_4$. But we may compute this cohomology group as the space of holomorphic $1$-forms on $M_0,4$ with at most logarithmic poles at infinity. Moreover, the cross-ratio $chi$ may be considered as a holomorphic function on $M_0,4$ and its logarithmic derivative $d log(chi)$ is such a 1-form with log poles. (Note that differentiating your analogue of the Bianchi identity gets rid of the $pi i$, so we really get something exactly like the usual Bianchi identity!)
answered 52 mins ago
Dan Petersen
24.4k268131
add a comment |Â
up vote
4
down vote
up vote
4
down vote
There is an indirect connection which goes via the representation theory of the symmetric group. The symmetries of the Riemann tensor are equivalent to saying that $R$ transforms according to the two-dimensional irreducible representation of $mathbb S_4$ corresponding to the partition $[2,2]$. On the other hand let us consider the moduli space $M_0,4$ parametrizing four distinct ordered points on the Riemann sphere up to Möbius transformations. The cohomology group $H^1(M_0,4,mathbf C)$ is $2$-dimensional, and transforms according to the representation $[2,2]$ under its natural action of $mathbb S_4$. But we may compute this cohomology group as the space of holomorphic $1$-forms on $M_0,4$ with at most logarithmic poles at infinity. Moreover, the cross-ratio $chi$ may be considered as a holomorphic function on $M_0,4$ and its logarithmic derivative $d log(chi)$ is such a 1-form with log poles. (Note that differentiating your analogue of the Bianchi identity gets rid of the $pi i$, so we really get something exactly like the usual Bianchi identity!)
answered 52 mins ago
Dan Petersen
24.4k268131
There is an indirect connection which goes via the representation theory of the symmetric group. The symmetries of the Riemann tensor are equivalent to saying that $R$ transforms according to the two-dimensional irreducible representation of $mathbb S_4$ corresponding to the partition $[2,2]$. On the other hand let us consider the moduli space $M_0,4$ parametrizing four distinct ordered points on the Riemann sphere up to Möbius transformations. The cohomology group $H^1(M_0,4,mathbf C)$ is $2$-dimensional, and transforms according to the representation $[2,2]$ under its natural action of $mathbb S_4$. But we may compute this cohomology group as the space of holomorphic $1$-forms on $M_0,4$ with at most logarithmic poles at infinity. Moreover, the cross-ratio $chi$ may be considered as a holomorphic function on $M_0,4$ and its logarithmic derivative $d log(chi)$ is such a 1-form with log poles. (Note that differentiating your analogue of the Bianchi identity gets rid of the $pi i$, so we really get something exactly like the usual Bianchi identity!)
answered 52 mins ago
Dan Petersen
24.4k268131
edited 22 mins ago
answered 52 mins ago
Dan Petersen
24.4k268131
answered 52 mins ago
Dan Petersen
24.4k268131
answered 52 mins ago
Dan Petersen
24.4k268131
24.4k268131
add a comment |Â
add a comment |Â
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