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How does one complexify a real $n$-dimensional Riemannian manifold $(M,g)$?
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If $V$ is a real vector space, then the complexification of $V$ is formally defined as $V^mathbbC=Votimes_mathbbRmathbbC$. Is there an analogous complexification operation for a real $n$-dimensional Riemannian manifold $(M,g)$?
Idea: The notion of complexification for Lie groups exists, so perhaps one can "complexify" a real Riemannian manifold by realizing it as a Lie group (or the quotient of one). It seems that under complexification of a real manifold some additional information must be added to determine a complex structure.
The reason I ask this is because I am looking through the Riemannian holonomy section of this article and it states that "the complexified holonomies $SO(n,mathbbC)$, $G_2(mathbbC)$, and $Spin(7,mathbbC)$ may be realized from complexifying real analytic Riemannian manifolds."
Any help would be much appreciated!
differential-geometry riemannian-geometry
asked Jul 24 at 15:52
Multivariablecalculus
495313
add a comment |Â
up vote
1
down vote
favorite
If $V$ is a real vector space, then the complexification of $V$ is formally defined as $V^mathbbC=Votimes_mathbbRmathbbC$. Is there an analogous complexification operation for a real $n$-dimensional Riemannian manifold $(M,g)$?
Idea: The notion of complexification for Lie groups exists, so perhaps one can "complexify" a real Riemannian manifold by realizing it as a Lie group (or the quotient of one). It seems that under complexification of a real manifold some additional information must be added to determine a complex structure.
The reason I ask this is because I am looking through the Riemannian holonomy section of this article and it states that "the complexified holonomies $SO(n,mathbbC)$, $G_2(mathbbC)$, and $Spin(7,mathbbC)$ may be realized from complexifying real analytic Riemannian manifolds."
Any help would be much appreciated!
differential-geometry riemannian-geometry
asked Jul 24 at 15:52
Multivariablecalculus
495313
1
Most manifolds are not quotients of Lie groups - how do you plan on complexifying those? What properties do you want your complexification to have?
– Jason DeVito
Jul 24 at 15:57
I would like to obtain a complexified holonomy for the holonomy group of $M$, e.g. if $Hol(M)=SO(n)$ then the complexified holonomy is $SO(n,mathbbC)$ which is realized by complexifiying the real analytic Riemannian manifold $M$. @JasonDeVito
– Multivariablecalculus
Jul 24 at 16:17
1
Hm, I'm afraid I'm useless, except to say that if the complexification is a manifold, then generically the holonomy group is all of $SO(n)$. So the complexification of a generic manifold will not be generic.
– Jason DeVito
Jul 24 at 16:32
1
I think things are basically hopeless except in the case of symmetric spaces. Google "complexification symmetric space" for some results.
– Qiaochu Yuan
Jul 24 at 17:09
The reason I ask is because I am looking through the Berger classification section of this article and it states that "the complexified holonomies $SO(n, mathbbC),$ $G_2(mathbbC),$ and $Spin(7,mathbbC)$ may be realized from complexifying real analytic Riemannian manifolds." @JasonDeVito
– Multivariablecalculus
Jul 24 at 17:37
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $V$ is a real vector space, then the complexification of $V$ is formally defined as $V^mathbbC=Votimes_mathbbRmathbbC$. Is there an analogous complexification operation for a real $n$-dimensional Riemannian manifold $(M,g)$?
Idea: The notion of complexification for Lie groups exists, so perhaps one can "complexify" a real Riemannian manifold by realizing it as a Lie group (or the quotient of one). It seems that under complexification of a real manifold some additional information must be added to determine a complex structure.
The reason I ask this is because I am looking through the Riemannian holonomy section of this article and it states that "the complexified holonomies $SO(n,mathbbC)$, $G_2(mathbbC)$, and $Spin(7,mathbbC)$ may be realized from complexifying real analytic Riemannian manifolds."
Any help would be much appreciated!
differential-geometry riemannian-geometry
asked Jul 24 at 15:52
Multivariablecalculus
495313
If $V$ is a real vector space, then the complexification of $V$ is formally defined as $V^mathbbC=Votimes_mathbbRmathbbC$. Is there an analogous complexification operation for a real $n$-dimensional Riemannian manifold $(M,g)$?
Idea: The notion of complexification for Lie groups exists, so perhaps one can "complexify" a real Riemannian manifold by realizing it as a Lie group (or the quotient of one). It seems that under complexification of a real manifold some additional information must be added to determine a complex structure.
The reason I ask this is because I am looking through the Riemannian holonomy section of this article and it states that "the complexified holonomies $SO(n,mathbbC)$, $G_2(mathbbC)$, and $Spin(7,mathbbC)$ may be realized from complexifying real analytic Riemannian manifolds."
Any help would be much appreciated!
differential-geometry riemannian-geometry
asked Jul 24 at 15:52
Multivariablecalculus
495313
edited Jul 24 at 19:44
asked Jul 24 at 15:52
Multivariablecalculus
495313
asked Jul 24 at 15:52
Multivariablecalculus
495313
asked Jul 24 at 15:52
Multivariablecalculus
495313
495313
1
Most manifolds are not quotients of Lie groups - how do you plan on complexifying those? What properties do you want your complexification to have?
– Jason DeVito
Jul 24 at 15:57
I would like to obtain a complexified holonomy for the holonomy group of $M$, e.g. if $Hol(M)=SO(n)$ then the complexified holonomy is $SO(n,mathbbC)$ which is realized by complexifiying the real analytic Riemannian manifold $M$. @JasonDeVito
– Multivariablecalculus
Jul 24 at 16:17
1
Hm, I'm afraid I'm useless, except to say that if the complexification is a manifold, then generically the holonomy group is all of $SO(n)$. So the complexification of a generic manifold will not be generic.
– Jason DeVito
Jul 24 at 16:32
1
I think things are basically hopeless except in the case of symmetric spaces. Google "complexification symmetric space" for some results.
– Qiaochu Yuan
Jul 24 at 17:09
The reason I ask is because I am looking through the Berger classification section of this article and it states that "the complexified holonomies $SO(n, mathbbC),$ $G_2(mathbbC),$ and $Spin(7,mathbbC)$ may be realized from complexifying real analytic Riemannian manifolds." @JasonDeVito
– Multivariablecalculus
Jul 24 at 17:37
add a comment |Â
1
Most manifolds are not quotients of Lie groups - how do you plan on complexifying those? What properties do you want your complexification to have?
– Jason DeVito
Jul 24 at 15:57
I would like to obtain a complexified holonomy for the holonomy group of $M$, e.g. if $Hol(M)=SO(n)$ then the complexified holonomy is $SO(n,mathbbC)$ which is realized by complexifiying the real analytic Riemannian manifold $M$. @JasonDeVito
– Multivariablecalculus
Jul 24 at 16:17
1
Hm, I'm afraid I'm useless, except to say that if the complexification is a manifold, then generically the holonomy group is all of $SO(n)$. So the complexification of a generic manifold will not be generic.
– Jason DeVito
Jul 24 at 16:32
1
I think things are basically hopeless except in the case of symmetric spaces. Google "complexification symmetric space" for some results.
– Qiaochu Yuan
Jul 24 at 17:09
The reason I ask is because I am looking through the Berger classification section of this article and it states that "the complexified holonomies $SO(n, mathbbC),$ $G_2(mathbbC),$ and $Spin(7,mathbbC)$ may be realized from complexifying real analytic Riemannian manifolds." @JasonDeVito
– Multivariablecalculus
Jul 24 at 17:37
1
1
Most manifolds are not quotients of Lie groups - how do you plan on complexifying those? What properties do you want your complexification to have?
– Jason DeVito
Jul 24 at 15:57
Most manifolds are not quotients of Lie groups - how do you plan on complexifying those? What properties do you want your complexification to have?
– Jason DeVito
Jul 24 at 15:57
I would like to obtain a complexified holonomy for the holonomy group of $M$, e.g. if $Hol(M)=SO(n)$ then the complexified holonomy is $SO(n,mathbbC)$ which is realized by complexifiying the real analytic Riemannian manifold $M$. @JasonDeVito
– Multivariablecalculus
Jul 24 at 16:17
I would like to obtain a complexified holonomy for the holonomy group of $M$, e.g. if $Hol(M)=SO(n)$ then the complexified holonomy is $SO(n,mathbbC)$ which is realized by complexifiying the real analytic Riemannian manifold $M$. @JasonDeVito
– Multivariablecalculus
Jul 24 at 16:17
1
1
Hm, I'm afraid I'm useless, except to say that if the complexification is a manifold, then generically the holonomy group is all of $SO(n)$. So the complexification of a generic manifold will not be generic.
– Jason DeVito
Jul 24 at 16:32
Hm, I'm afraid I'm useless, except to say that if the complexification is a manifold, then generically the holonomy group is all of $SO(n)$. So the complexification of a generic manifold will not be generic.
– Jason DeVito
Jul 24 at 16:32
1
1
I think things are basically hopeless except in the case of symmetric spaces. Google "complexification symmetric space" for some results.
– Qiaochu Yuan
Jul 24 at 17:09
I think things are basically hopeless except in the case of symmetric spaces. Google "complexification symmetric space" for some results.
– Qiaochu Yuan
Jul 24 at 17:09
The reason I ask is because I am looking through the Berger classification section of this article and it states that "the complexified holonomies $SO(n, mathbbC),$ $G_2(mathbbC),$ and $Spin(7,mathbbC)$ may be realized from complexifying real analytic Riemannian manifolds." @JasonDeVito
– Multivariablecalculus
Jul 24 at 17:37
The reason I ask is because I am looking through the Berger classification section of this article and it states that "the complexified holonomies $SO(n, mathbbC),$ $G_2(mathbbC),$ and $Spin(7,mathbbC)$ may be realized from complexifying real analytic Riemannian manifolds." @JasonDeVito
– Multivariablecalculus
Jul 24 at 17:37
add a comment |Â
1 Answer
1
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up vote
0
down vote
accepted
The following is by M.G.:
I believe the following is meant:
Every smooth (real) manifold $M$ has a (unique) real-analytic structure compatible with the smooth structure. So, cover $M$ with real-analytic charts, i.e. whose transition functions are real-analytic diffeomorphisms
$$
phi_ij:=phi_j^-1circphi_i: U_ij:=phi_i^-1(phi_i(U_i)capphi_j(U_j))to U_ji
$$
One can find open subsets $U_i^mathbbCsubseteqmathbbC^n$ with $U_i^mathbbCcapmathbbR^n=U_i$ and $U_ij^mathbbCcapmathbbR^n=U_ij$ such that the (real-analytic) $phi_ij$ extend to biholomorphisms $phi_ij^mathbbC:U_ij^mathbbCto U_ji^mathbbC$ satisfying the usual cocycle conditions. Then the complexification $M^mathbbC$ is defined as a quotient space of the disjoint union, $left(coprod_i U_i^mathbbCright)/sim$, where $z_isim z_j$ iff $z_iin U_ij^mathbbC$ and $z_j = phi_ij^mathbbC(z_i)$ (this works because of the cocycle conditions). The maps $U_i^mathbbChookrightarrowcoprod U_i^mathbbC$ induce coordinate charts $U_i^mathbbCto M^mathbbC$ with biholomorphic transition functions.
This and the details around it are part (of the proof of) Bruhat-Whitney's theorem* on the existence of $M^mathbbC$. Moreover, complexification is functorial in the obvious way. By Grauert, $M^mathbbC$ is in fact a Stein manifold.
*F. Bruhat and H. Whitney, Quelques propriétés fondamentales des ensembles analytiques-réels, Comment. Math. Helv. 33, 132-160 (1959).
answered Jul 24 at 21:49
Multivariablecalculus
495313
Is the complexification so obtained unique in some way?
– Alex M.
Aug 2 at 16:50
2
@AlexM. sorry for replying so late, I didn't get a notification and saw your comment accidentally by following the link from MO. Yes, there is uniqueness of the germ of $M$ in the following sense (btw, also part of Bruhat-Whitney's statement): if $N_1$ and $N_2$ are two complex manifolds containing $M$ as real-analytic and totally real submanifold with $dim_mathbbCN_1=dim_mathbbCN_2=dim_mathbbRM$, then some neighbourhoods of $M$ in $N_1$ and $N_2$ are biholomorphic.
– M.G.
Aug 6 at 20:38
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The following is by M.G.:
I believe the following is meant:
Every smooth (real) manifold $M$ has a (unique) real-analytic structure compatible with the smooth structure. So, cover $M$ with real-analytic charts, i.e. whose transition functions are real-analytic diffeomorphisms
$$
phi_ij:=phi_j^-1circphi_i: U_ij:=phi_i^-1(phi_i(U_i)capphi_j(U_j))to U_ji
$$
One can find open subsets $U_i^mathbbCsubseteqmathbbC^n$ with $U_i^mathbbCcapmathbbR^n=U_i$ and $U_ij^mathbbCcapmathbbR^n=U_ij$ such that the (real-analytic) $phi_ij$ extend to biholomorphisms $phi_ij^mathbbC:U_ij^mathbbCto U_ji^mathbbC$ satisfying the usual cocycle conditions. Then the complexification $M^mathbbC$ is defined as a quotient space of the disjoint union, $left(coprod_i U_i^mathbbCright)/sim$, where $z_isim z_j$ iff $z_iin U_ij^mathbbC$ and $z_j = phi_ij^mathbbC(z_i)$ (this works because of the cocycle conditions). The maps $U_i^mathbbChookrightarrowcoprod U_i^mathbbC$ induce coordinate charts $U_i^mathbbCto M^mathbbC$ with biholomorphic transition functions.
This and the details around it are part (of the proof of) Bruhat-Whitney's theorem* on the existence of $M^mathbbC$. Moreover, complexification is functorial in the obvious way. By Grauert, $M^mathbbC$ is in fact a Stein manifold.
*F. Bruhat and H. Whitney, Quelques propriétés fondamentales des ensembles analytiques-réels, Comment. Math. Helv. 33, 132-160 (1959).
answered Jul 24 at 21:49
Multivariablecalculus
495313
Is the complexification so obtained unique in some way?
– Alex M.
Aug 2 at 16:50
2
@AlexM. sorry for replying so late, I didn't get a notification and saw your comment accidentally by following the link from MO. Yes, there is uniqueness of the germ of $M$ in the following sense (btw, also part of Bruhat-Whitney's statement): if $N_1$ and $N_2$ are two complex manifolds containing $M$ as real-analytic and totally real submanifold with $dim_mathbbCN_1=dim_mathbbCN_2=dim_mathbbRM$, then some neighbourhoods of $M$ in $N_1$ and $N_2$ are biholomorphic.
– M.G.
Aug 6 at 20:38
add a comment |Â
up vote
0
down vote
accepted
The following is by M.G.:
I believe the following is meant:
Every smooth (real) manifold $M$ has a (unique) real-analytic structure compatible with the smooth structure. So, cover $M$ with real-analytic charts, i.e. whose transition functions are real-analytic diffeomorphisms
$$
phi_ij:=phi_j^-1circphi_i: U_ij:=phi_i^-1(phi_i(U_i)capphi_j(U_j))to U_ji
$$
One can find open subsets $U_i^mathbbCsubseteqmathbbC^n$ with $U_i^mathbbCcapmathbbR^n=U_i$ and $U_ij^mathbbCcapmathbbR^n=U_ij$ such that the (real-analytic) $phi_ij$ extend to biholomorphisms $phi_ij^mathbbC:U_ij^mathbbCto U_ji^mathbbC$ satisfying the usual cocycle conditions. Then the complexification $M^mathbbC$ is defined as a quotient space of the disjoint union, $left(coprod_i U_i^mathbbCright)/sim$, where $z_isim z_j$ iff $z_iin U_ij^mathbbC$ and $z_j = phi_ij^mathbbC(z_i)$ (this works because of the cocycle conditions). The maps $U_i^mathbbChookrightarrowcoprod U_i^mathbbC$ induce coordinate charts $U_i^mathbbCto M^mathbbC$ with biholomorphic transition functions.
This and the details around it are part (of the proof of) Bruhat-Whitney's theorem* on the existence of $M^mathbbC$. Moreover, complexification is functorial in the obvious way. By Grauert, $M^mathbbC$ is in fact a Stein manifold.
*F. Bruhat and H. Whitney, Quelques propriétés fondamentales des ensembles analytiques-réels, Comment. Math. Helv. 33, 132-160 (1959).
answered Jul 24 at 21:49
Multivariablecalculus
495313
Is the complexification so obtained unique in some way?
– Alex M.
Aug 2 at 16:50
2
@AlexM. sorry for replying so late, I didn't get a notification and saw your comment accidentally by following the link from MO. Yes, there is uniqueness of the germ of $M$ in the following sense (btw, also part of Bruhat-Whitney's statement): if $N_1$ and $N_2$ are two complex manifolds containing $M$ as real-analytic and totally real submanifold with $dim_mathbbCN_1=dim_mathbbCN_2=dim_mathbbRM$, then some neighbourhoods of $M$ in $N_1$ and $N_2$ are biholomorphic.
– M.G.
Aug 6 at 20:38
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The following is by M.G.:
I believe the following is meant:
Every smooth (real) manifold $M$ has a (unique) real-analytic structure compatible with the smooth structure. So, cover $M$ with real-analytic charts, i.e. whose transition functions are real-analytic diffeomorphisms
$$
phi_ij:=phi_j^-1circphi_i: U_ij:=phi_i^-1(phi_i(U_i)capphi_j(U_j))to U_ji
$$
One can find open subsets $U_i^mathbbCsubseteqmathbbC^n$ with $U_i^mathbbCcapmathbbR^n=U_i$ and $U_ij^mathbbCcapmathbbR^n=U_ij$ such that the (real-analytic) $phi_ij$ extend to biholomorphisms $phi_ij^mathbbC:U_ij^mathbbCto U_ji^mathbbC$ satisfying the usual cocycle conditions. Then the complexification $M^mathbbC$ is defined as a quotient space of the disjoint union, $left(coprod_i U_i^mathbbCright)/sim$, where $z_isim z_j$ iff $z_iin U_ij^mathbbC$ and $z_j = phi_ij^mathbbC(z_i)$ (this works because of the cocycle conditions). The maps $U_i^mathbbChookrightarrowcoprod U_i^mathbbC$ induce coordinate charts $U_i^mathbbCto M^mathbbC$ with biholomorphic transition functions.
This and the details around it are part (of the proof of) Bruhat-Whitney's theorem* on the existence of $M^mathbbC$. Moreover, complexification is functorial in the obvious way. By Grauert, $M^mathbbC$ is in fact a Stein manifold.
*F. Bruhat and H. Whitney, Quelques propriétés fondamentales des ensembles analytiques-réels, Comment. Math. Helv. 33, 132-160 (1959).
answered Jul 24 at 21:49
Multivariablecalculus
495313
The following is by M.G.:
I believe the following is meant:
Every smooth (real) manifold $M$ has a (unique) real-analytic structure compatible with the smooth structure. So, cover $M$ with real-analytic charts, i.e. whose transition functions are real-analytic diffeomorphisms
$$
phi_ij:=phi_j^-1circphi_i: U_ij:=phi_i^-1(phi_i(U_i)capphi_j(U_j))to U_ji
$$
One can find open subsets $U_i^mathbbCsubseteqmathbbC^n$ with $U_i^mathbbCcapmathbbR^n=U_i$ and $U_ij^mathbbCcapmathbbR^n=U_ij$ such that the (real-analytic) $phi_ij$ extend to biholomorphisms $phi_ij^mathbbC:U_ij^mathbbCto U_ji^mathbbC$ satisfying the usual cocycle conditions. Then the complexification $M^mathbbC$ is defined as a quotient space of the disjoint union, $left(coprod_i U_i^mathbbCright)/sim$, where $z_isim z_j$ iff $z_iin U_ij^mathbbC$ and $z_j = phi_ij^mathbbC(z_i)$ (this works because of the cocycle conditions). The maps $U_i^mathbbChookrightarrowcoprod U_i^mathbbC$ induce coordinate charts $U_i^mathbbCto M^mathbbC$ with biholomorphic transition functions.
This and the details around it are part (of the proof of) Bruhat-Whitney's theorem* on the existence of $M^mathbbC$. Moreover, complexification is functorial in the obvious way. By Grauert, $M^mathbbC$ is in fact a Stein manifold.
*F. Bruhat and H. Whitney, Quelques propriétés fondamentales des ensembles analytiques-réels, Comment. Math. Helv. 33, 132-160 (1959).
answered Jul 24 at 21:49
Multivariablecalculus
495313
answered Jul 24 at 21:49
Multivariablecalculus
495313
answered Jul 24 at 21:49
Multivariablecalculus
495313
answered Jul 24 at 21:49
Multivariablecalculus
495313
495313
Is the complexification so obtained unique in some way?
– Alex M.
Aug 2 at 16:50
2
@AlexM. sorry for replying so late, I didn't get a notification and saw your comment accidentally by following the link from MO. Yes, there is uniqueness of the germ of $M$ in the following sense (btw, also part of Bruhat-Whitney's statement): if $N_1$ and $N_2$ are two complex manifolds containing $M$ as real-analytic and totally real submanifold with $dim_mathbbCN_1=dim_mathbbCN_2=dim_mathbbRM$, then some neighbourhoods of $M$ in $N_1$ and $N_2$ are biholomorphic.
– M.G.
Aug 6 at 20:38
add a comment |Â
Is the complexification so obtained unique in some way?
– Alex M.
Aug 2 at 16:50
2
@AlexM. sorry for replying so late, I didn't get a notification and saw your comment accidentally by following the link from MO. Yes, there is uniqueness of the germ of $M$ in the following sense (btw, also part of Bruhat-Whitney's statement): if $N_1$ and $N_2$ are two complex manifolds containing $M$ as real-analytic and totally real submanifold with $dim_mathbbCN_1=dim_mathbbCN_2=dim_mathbbRM$, then some neighbourhoods of $M$ in $N_1$ and $N_2$ are biholomorphic.
– M.G.
Aug 6 at 20:38
Is the complexification so obtained unique in some way?
– Alex M.
Aug 2 at 16:50
Is the complexification so obtained unique in some way?
– Alex M.
Aug 2 at 16:50
2
2
@AlexM. sorry for replying so late, I didn't get a notification and saw your comment accidentally by following the link from MO. Yes, there is uniqueness of the germ of $M$ in the following sense (btw, also part of Bruhat-Whitney's statement): if $N_1$ and $N_2$ are two complex manifolds containing $M$ as real-analytic and totally real submanifold with $dim_mathbbCN_1=dim_mathbbCN_2=dim_mathbbRM$, then some neighbourhoods of $M$ in $N_1$ and $N_2$ are biholomorphic.
– M.G.
Aug 6 at 20:38
@AlexM. sorry for replying so late, I didn't get a notification and saw your comment accidentally by following the link from MO. Yes, there is uniqueness of the germ of $M$ in the following sense (btw, also part of Bruhat-Whitney's statement): if $N_1$ and $N_2$ are two complex manifolds containing $M$ as real-analytic and totally real submanifold with $dim_mathbbCN_1=dim_mathbbCN_2=dim_mathbbRM$, then some neighbourhoods of $M$ in $N_1$ and $N_2$ are biholomorphic.
– M.G.
Aug 6 at 20:38
add a comment |Â
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1
Most manifolds are not quotients of Lie groups - how do you plan on complexifying those? What properties do you want your complexification to have?
– Jason DeVito
Jul 24 at 15:57
I would like to obtain a complexified holonomy for the holonomy group of $M$, e.g. if $Hol(M)=SO(n)$ then the complexified holonomy is $SO(n,mathbbC)$ which is realized by complexifiying the real analytic Riemannian manifold $M$. @JasonDeVito
– Multivariablecalculus
Jul 24 at 16:17
1
Hm, I'm afraid I'm useless, except to say that if the complexification is a manifold, then generically the holonomy group is all of $SO(n)$. So the complexification of a generic manifold will not be generic.
– Jason DeVito
Jul 24 at 16:32
1
I think things are basically hopeless except in the case of symmetric spaces. Google "complexification symmetric space" for some results.
– Qiaochu Yuan
Jul 24 at 17:09
The reason I ask is because I am looking through the Berger classification section of this article and it states that "the complexified holonomies $SO(n, mathbbC),$ $G_2(mathbbC),$ and $Spin(7,mathbbC)$ may be realized from complexifying real analytic Riemannian manifolds." @JasonDeVito
– Multivariablecalculus
Jul 24 at 17:37