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The homogeneous space $M=G/K=SO(2n+1)/U(n)$ is not a symmetric space but a generalized flag manifold of the compact simple Lie group $SO(2n+1)$. It arises from the more general family $$ M_{n, p}:=SO(2n+1)/(U(p) \times SO(2(n-p)+1) ) $$ for $p=n$.

Notice that for any $2\leq p\leq n$ the isotropy representation of $M_{n, p}$ decomposes intro two isotropy summands: $${\frak m}\cong T_{o}M_{n, p}=\frak{m}_{1}\oplus\frak{m}_{2}.$$ This is true, since in this case you paint black in the Dynkin diagram of $G=SO(2n+1)$ a simple root of Dynkin mark equal to 2. It is easy to see that $[\frak{m}, \frak{m}]\neq \frak{k}$, in particular one computes $$ [\frak{m}_{1}, \frak{m}_{1}]\subset\frak{k}\oplus\frak{m}_{2}, \quad [\frak{m}_{1}, \frak{m}_{2}]\subset\frak{m}_{1},\quad [\frak{m}_{2}, \frak{m}_{2}]\subset\frak{k} $$ For $p=1$ one gets the Hermitian symmetric space $SO(2n+1)/SO(2)\times SO(2n-1)$, which of course is isotropy irreducible (in this case, in the Dynkin diagram of $G=SO(2n+1)$ we paint black the first simple root, which has Dynkin mark equal to 1, recall that the highest root of $SO(2n+1)$ is given by $\tilde{\alpha}=\alpha_1+2\alpha_2+\cdots+2\alpha_{n}$ where $\{\alpha_1, \ldots, \alpha_{n}\}$ is a basis of simple roots. The coefficients $\{1, 2, \ldots, 2\}$ are the so called Dynkin marks).

In fact, the only generalized flag manifolds which are the same time symmetric spaces are the (compact) Hermitian symmetric spaces and these are the unique flag manifolds which are isotropy irreducible. (see Which Kahler Manifolds are also Einstein Manifolds? for more details)

Finally notice that a smooth manifold can has more than one expressions as a homogeneous space, $M=G/K=G'/K'$. For example $$S^{6}=SO(7)/SO(6)=G_2/SU(3), \quad S^{7}=SO(8)/SO(7)=Sp(2)/Sp(1)=Spin(7)/G_2.$$ However, only the pairs $(SO(7), SO(6))$ and $(SO(8), SO(7))$ are symmetric pairs (and so passing to the double coverings you get simply-connected symmetric spaces). And similar in your example: the first complex projective space $$SO(5)/U(2)=ℂP^{3}_{{\frak m}_{1}\oplus{\frak m}_{2}}$$ is not a symmetric space but a flag manifold with two isotropy summands, but $SU(4)/U(3)=ℂP^{3}$ is a Hermitian symmetric space (irreducible).

The homogeneous space $M=G/K=SO(2n+1)/U(n)$ is not a symmetric space but a generalized flag manifold of the compact simple Lie group $SO(2n+1)$. It arises from the more general family $$ M_{n, p}:=SO(2n+1)/(U(p) \times SO(2(n-p)+1) ) $$ for $p=n$.

Notice that for any $2\leq p\leq n$ the isotropy representation of $M_{n, p}$ decomposes intro two isotropy summands: $${\frak m}\cong T_{o}M_{n, p}=\frak{m}_{1}\oplus\frak{m}_{2}.$$ This is true, since in this case you paint black in the Dynkin diagram of $G=SO(2n+1)$ a simple root of Dynkin mark equal to 2. It is easy to see that $[\frak{m}, \frak{m}]\neq \frak{k}$, in particular one computes $$ [\frak{m}_{1}, \frak{m}_{1}]\subset\frak{k}\oplus\frak{m}_{2}, \quad [\frak{m}_{1}, \frak{m}_{2}]\subset\frak{m}_{1},\quad [\frak{m}_{2}, \frak{m}_{2}]\subset\frak{k} $$ For $p=1$ one gets the Hermitian symmetric space $SO(2n+1)/SO(2)\times SO(2n-1)$, which of course is isotropy irreducible (in this case, in the Dynkin diagram of $G=SO(2n+1)$ we paint black the first simple root, which has Dynkin mark equal to 1, recall that the highest root of $SO(2n+1)$ is given by $\tilde{\alpha}=\alpha_1+2\alpha_2+\cdots+2\alpha_{n}$ where $\{\alpha_1, \ldots, \alpha_{n}\}$ is a basis of simple roots. The coefficients $\{1, 2, \ldots, 2\}$ are the so called Dynkin marks).

In fact, the only generalized flag manifolds which are the same time symmetric spaces are the (compact) Hermitian symmetric spaces and these are the unique flag manifolds which are isotropy irreducible. (see Which Kahler Manifolds are also Einstein Manifolds? for more details)

Finally notice that a smooth manifold can has more than one expressions as a homogeneous space, $M=G/K=G'/K'$. For example $$S^{6}=SO(7)/SO(6)=G_2/SU(3), \quad S^{7}=SO(8)/SO(7)=Sp(2)/Sp(1)=Spin(7)/G_2.$$ However, only the pairs $(SO(7), SO(6))$ and $(SO(8), SO(7))$ are symmetric pairs (and so passing to the double coverings you get simply-connected symmetric spaces). And similar in your example: the first complex projective space $$SO(5)/U(2)=ℂP^{3}_{{\frak m}_{1}\oplus{\frak m}_{2}}$$ is not a symmetric space but a flag manifold with two isotropy summands, but $SU(4)/U(3)=ℂP^{3}$ is a Hermitian symmetric space (irreducible).

The homogeneous space $M=G/K=SO(2n+1)/U(n)$ is not a symmetric space but a generalized flag manifold of the compact simple Lie group $SO(2n+1)$. It arises from the more general family $$ M_{n, p}:=SO(2n+1)/(U(p) \times SO(2(n-p)+1) ) $$ for $p=n$.

Notice that for any $2\leq p\leq n$ the isotropy representation of $M_{n, p}$ decomposes intro two isotropy summands: $${\frak m}\cong T_{o}M_{n, p}=\frak{m}_{1}\oplus\frak{m}_{2}.$$ This is true, since in this case you paint black in the Dynkin diagram of $G=SO(2n+1)$ a simple root of Dynkin mark equal to 2. It is easy to see that $[\frak{m}, \frak{m}]\neq \frak{k}$, in particular one computes $$ [\frak{m}_{1}, \frak{m}_{1}]\subset\frak{k}\oplus\frak{m}_{2}, \quad [\frak{m}_{1}, \frak{m}_{2}]\subset\frak{m}_{1},\quad [\frak{m}_{2}, \frak{m}_{2}]\subset\frak{k} $$ For $p=1$ one gets the Hermitian symmetric space $SO(2n+1)/SO(2)\times SO(2n-1)$, which of course is isotropy irreducible (in this case, in the Dynkin diagram of $G=SO(2n+1)$ we paint black the first simple root, which has Dynkin mark equal to 1, recall that the highest root of $SO(2n+1)$ is given by $\tilde{\alpha}=\alpha_1+2\alpha_2+\cdots+2\alpha_{n}$ where $\{\alpha_1, \ldots, \alpha_{n}\}$ is a basis of simple roots. The coefficients $\{1, 2, \ldots, 2\}$ are the so called Dynkin marks).

In fact, the only generalized flag manifolds which are the same time symmetric spaces are the (compact) Hermitian symmetric spaces and these are the unique flag manifolds which are isotropy irreducible. (see Which Kahler Manifolds are also Einstein Manifolds? for more details)

Finally notice that a smooth manifold can has more than one expressions as a homogeneous space, $M=G/K=G'/K'$. For example $$S^{6}=SO(7)/SO(6)=G_2/SU(3), \quad S^{7}=SO(8)/SO(7)=Sp(2)/Sp(1)=Spin(7)/G_2.$$ However, only the pairs $(SO(7), SO(6))$ and $(SO(8), SO(7))$ are symmetric pairs (and so passing to the double coverings you get simply-connected symmetric spaces). And similar in your example: the first complex projective space $SO(5)/U(2)=ℂP^{3}_{{\frak m}_{1}\oplus{\frak m}_{2}}$ is not a symmetric space but a flag manifold with two isotropy summands, but $SU(4)/U(3)=ℂP^{3}$ is a Hermitian symmetric space (irreducible).

The homogeneous space $M=G/K=SO(2n+1)/U(n)$ is not a symmetric space but a generalized flag manifold of the compact simple Lie group $SO(2n+1)$. It arises from the more general family $$ M_{n, p}:=SO(2n+1)/(U(p) \times SO(2(n-p)+1) ) $$ for $p=n$.

Notice that for any $2\leq p\leq n$ the isotropy representation of $M_{n, p}$ decomposes intro two isotropy summands: $${\frak m}\cong T_{o}M_{n, p}=\frak{m}_{1}\oplus\frak{m}_{2}.$$ This is true, since in this case you paint black in the Dynkin diagram of $G=SO(2n+1)$ a simple root of Dynkin mark equal to 2. It is easy to see that $[\frak{m}, \frak{m}]\neq \frak{k}$, in particular one computes $$ [\frak{m}_{1}, \frak{m}_{1}]\subset\frak{k}\oplus\frak{m}_{2}, \quad [\frak{m}_{1}, \frak{m}_{2}]\subset\frak{m}_{1},\quad [\frak{m}_{2}, \frak{m}_{2}]\subset\frak{k} $$ For $p=1$ one gets the Hermitian symmetric space $SO(2n+1)/SO(2)\times SO(2n-1)$, which of course is isotropy irreducible (in this case, in the Dynkin diagram of $G=SO(2n+1)$ we paint black the first simple root, which has Dynkin mark equal to 1, recall that the highest root of $SO(2n+1)$ is given by $\tilde{\alpha}=\alpha_1+2\alpha_2+\cdots+2\alpha_{n}$ where $\{\alpha_1, \ldots, \alpha_{n}\}$ is a basis of simple roots. The coefficients $\{1, 2, \ldots, 2\}$ are the so called Dynkin marks).

In fact, the only generalized flag manifolds which are the same time symmetric spaces are the (compact) Hermitian symmetric spaces and these are the unique flag manifolds which are isotropy irreducible. (see Which Kahler Manifolds are also Einstein Manifolds? for more details)

Finally notice that a smooth manifold can has more than one expressions as a homogeneous space, $M=G/K=G'/K'$. For example $$S^{6}=SO(7)/SO(6)=G_2/SU(3), \quad S^{7}=SO(8)/SO(7)=Sp(2)/Sp(1)=Spin(7)/G_2.$$ However, only the pairs $(SO(7), SO(6))$ and $(SO(8), SO(7))$ are symmetric pairs (and so passing to the double coverings you get simply-connected symmetric spaces). And similar in your example: the first complex projective space $$SO(5)/U(2)=ℂP^{3}_{{\frak m}_{1}\oplus{\frak m}_{2}}$$ is not a symmetric space but a flag manifold with two isotropy summands, but $SU(4)/U(3)=ℂP^{3}$ is a Hermitian symmetric space (irreducible).

The homogeneous space $M=G/K=SO(2n+1)/U(n)$ is not a symmetric space but a generalized flag manifold of the compact simple Lie group $SO(2n+1)$. It arises from the more general family $$ M_{n, p}:=SO(2n+1)/(U(p) \times SO(2(n-p)+1) ) $$ for $p=n$.

Notice that for any $2\leq p\leq n$ the isotropy representation of $M_{n, p}$ decomposes intro two isotropy summands, $${\frak m}\cong T_{o}M=\frak{m}_{1}\oplus\frak{m}_{2}.$$ For $p=1$ one gets the Hermitian symmetric space $SO(2n+1)/SO(2)\times SO(2n-1)$, which of course is isotropy irreducible. In fact, the only generalized flag manifolds which are the same time symmetric spaces are the (compact) Hermitian symmetric spaces and these are the unique flag manifolds which are isotropy irreducible. (see Which Kahler Manifolds are also Einstein Manifolds? for more details)

Finally notice that a smooth manifold can has more than one expressions as a homogeneous space, $M=G/K=G'/K'$. For example $$S^{6}=SO(7)/SO(6)=G_2/SU(3), \quad S^{7}=SO(8)/SO(7)=Sp(2)/Sp(1)=Spin(7)/G_2.$$ However, only the pairs $(SO(7), SO(6))$ and $(SO(8), SO(7))$ are symmetric pairs (and so passing to the double coverings you get simply-connected symmetric spaces). And similar in your example: the first complex projective space $SO(5)/U(2)=ℂP^{3}_{{\frak m}_{1}\oplus{\frak m}_{2}}$ is not a symmetric space but a flag manifold with two isotropy summands, but $SU(4)/U(3)=ℂP^{3}$ is a Hermitian symmetric space (irreducible).

The homogeneous space $M=G/K=SO(2n+1)/U(n)$ is not a symmetric space but a generalized flag manifold of the compact simple Lie group $SO(2n+1)$. It arises from the more general family $$ M_{n, p}:=SO(2n+1)/(U(p) \times SO(2(n-p)+1) ) $$ for $p=n$.

Notice that for any $2\leq p\leq n$ the isotropy representation of $M_{n, p}$ decomposes intro two isotropy summands: $${\frak m}\cong T_{o}M_{n, p}=\frak{m}_{1}\oplus\frak{m}_{2}.$$ This is true, since in this case you paint black in the Dynkin diagram of $G=SO(2n+1)$ a simple root of Dynkin mark equal to 2. It is easy to see that $[\frak{m}, \frak{m}]\neq \frak{k}$, in particular one computes $$ [\frak{m}_{1}, \frak{m}_{1}]\subset\frak{k}\oplus\frak{m}_{2}, \quad [\frak{m}_{1}, \frak{m}_{2}]\subset\frak{m}_{1},\quad [\frak{m}_{2}, \frak{m}_{2}]\subset\frak{k} $$ For $p=1$ one gets the Hermitian symmetric space $SO(2n+1)/SO(2)\times SO(2n-1)$, which of course is isotropy irreducible (in this case, in the Dynkin diagram of $G=SO(2n+1)$ we paint black the first simple root, which has Dynkin mark equal to 1, recall that the highest root of $SO(2n+1)$ is given by $\tilde{\alpha}=\alpha_1+2\alpha_2+\cdots+2\alpha_{n}$ where $\{\alpha_1, \ldots, \alpha_{n}\}$ is a basis of simple roots. The coefficients $\{1, 2, \ldots, 2\}$ are the so called Dynkin marks). 

In fact, the only generalized flag manifolds which are the same time symmetric spaces are the (compact) Hermitian symmetric spaces and these are the unique flag manifolds which are isotropy irreducible. (see Which Kahler Manifolds are also Einstein Manifolds? for more details)

Finally notice that a smooth manifold can has more than one expressions as a homogeneous space, $M=G/K=G'/K'$. For example $$S^{6}=SO(7)/SO(6)=G_2/SU(3), \quad S^{7}=SO(8)/SO(7)=Sp(2)/Sp(1)=Spin(7)/G_2.$$ However, only the pairs $(SO(7), SO(6))$ and $(SO(8), SO(7))$ are symmetric pairs (and so passing to the double coverings you get simply-connected symmetric spaces). And similar in your example: the first complex projective space $SO(5)/U(2)=ℂP^{3}_{{\frak m}_{1}\oplus{\frak m}_{2}}$ is not a symmetric space but a flag manifold with two isotropy summands, but $SU(4)/U(3)=ℂP^{3}$ is a Hermitian symmetric space (irreducible).