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Preamble
RIEMANNIAN SPACES WITH UNUSUAL HOLONOMY GROUPS
D. V. Alekseevsky
Functional Analysis and Its Applications, Vol. 2, No. 2, 1968, pp. 1–10.
Introduction
The problem of classifying the holonomy groups of Riemannian spaces was posed by Cartan and essentially solved by Berger \cite{1}. Berger provided a list of groups that can serve as the restricted holonomy group of a simply connected irreducible Riemannian space that is not locally symmetric. This list includes:
1. $SO(n)$
2. $U(n)$
3. $SU(n)$
4. $Sp(n)$
5. $Sp(n) \cdot Sp(1)$
6. $Spin(7)$ (for $n=8$)
7. $Spin(9)$ (for $n=16$)
8. $G_2$ (for $n=7$)
For a long time, it remained unknown whether all groups on this list could actually be realized as holonomy groups of Riemannian spaces. It was well known that the first four cases correspond to general Riemannian, Kähler, Ricci-flat Kähler, and hyper-Kähler manifolds, respectively. However, the existence of manifolds with holonomy groups $Sp(n) \cdot Sp(1)$, $Spin(7)$, $Spin(9)$, and $G_2$ remained an open question.
In this paper, we demonstrate that the groups $Spin(9)$ and $Spin(7)$ cannot be the holonomy groups of any Riemannian space that is not locally symmetric. Specifically, we prove that any Riemannian manifold with a holonomy group contained in $Spin(9)$ (for $n=16$) or $Spin(7)$ (for $n=8$) must be locally symmetric. Since the symmetric spaces with these holonomy groups are well-known (the Cayley projective plane $F_4/Spin(9)$ and its non-compact dual), this completes the classification for these cases.
1. Basic Definitions and Notation
Let $M$ be a Riemannian manifold of dimension $n$ with metric $g$. Let $\nabla$ denote the Levi-Civita connection and $R$ the curvature tensor. The restricted holonomy group $\Phi$ at a point $p \in M$ is a subgroup of $SO(n)$
Introduction
Generally, the holonomy group of an $n$-dimensional orientable Riemannian space is the special orthogonal group $SO(n)$ (this can be achieved by a small deformation of the metric in the neighborhood of an arbitrary point). When the holonomy group of a space differs from $SO(n)$, it carries significant information regarding the differential geometry and topology of that space. For instance, knowing the holonomy group of a symmetric space allows for the efficient calculation of its cohomology algebra (which, in this case, is isomorphic to the algebra of parallel differential forms) and enables the identification of the space up to its dual symmetric space. As shown by J. Hano and H. Ozeki, every connected linear Lie group can be realized as the holonomy group of a space with linear connection. For the holonomy groups of Riemannian spaces, however, the situation is different. Specifically, M. Berger proved the following theorem in 1953: The restricted holonomy group of a locally irreducible Riemannian space $M^n$ that is not locally symmetric is contained in the following list of compact linear Lie groups:
$SO(n)$, $U(n/2)$, $SU(n/2)$, $Sp(1) \cdot Sp(n/4)$, $Sp(n/4)$, $G_2$ ($n=7$), $Spin(7)$ ($n=8$), $Spin(9)$ ($n=16$). In the present article, we describe the curvature tensors of Riemannian spaces with various holonomy groups from Berger's list and derive several consequences. Let the tangent space to the Riemannian space at an arbitrary point be identified with its dual space via the Riemannian metric. Then the holonomy algebra $\Gamma$ of the space at point $p$ (i.e., the Lie algebra of the holonomy group) is identified with a subspace of $\Lambda^2 V = \{x \wedge y = x \otimes y - y \otimes x \mid x, y \in V\}$, and the curvature tensor of the space $M^n$ at that point is identified with an element of the space $S^2(\Gamma) \subset \Gamma \otimes \Gamma = \{A \otimes B + B \otimes A \mid A, B \in \Gamma\}$ that satisfies the Bianchi identity. The set $R(\Gamma)$ of tensors from $S^2(\Gamma)$ satisfying the Bianchi identity is called the space of curvature tensors of type $\Gamma$. This article describes the spaces $R(\Gamma)$ for the Lie algebras $\Gamma$ corresponding to the various Lie groups in Berger's list. For $n=4$, such a description is well known and, as shown by Goldberg and Kerr \cite{4}, leads to A. Z. Petrov's classification of gravitational fields in general relativity. The method of description is based on the following observation: the space $R(\Gamma)$ is invariant under the action of the Lie algebra $\Gamma$ on the space $S^2(\Gamma)$ induced by the representation of $\Gamma$ on $V$.
$R(\Gamma)$, $S^2(\Gamma)$, $so(3)$, $so(4)$, $R(2,2)$
$SO(n)$, $n > 4$
$U(m)$, $n = 2m$
$SU(m)$, $n = 2m$
Sp(l) Sp(m)4^f'
$Sp(m)$, $n = 4m$
$Spin(9)$, $n = 16$
Spin (7), n = S
$G_2, n = 7$
[TABLE:4] $(n-3)(n+2)/2, \dots, (n-3)$
$\Phi_0(SU(m)) = \mathcal{R}_0(U(n))$
$- (2m+3)(2m+1)m$
$\Phi_{(4, 0, \dots, 0), 0}(Sp(n)) = \Phi_0(Sp(1) + Sp(n))$
The adjoint representation of the algebra $\Gamma$ acts on $\Gamma$. Therefore, to describe the space, it is sufficient to decompose $S^2(\Gamma)$ into $\Gamma$-irreducible non-equivalent components and verify which of them satisfy the Bianchi identity. The main results obtained are summarized in Tables 1 and 2. The following notation is adopted in the tables: $\mathcal{R}(\Gamma)$ denotes the space of $\Gamma$-invariant tensors; $\mathcal{R}'(\Gamma)$ denotes the space of tensors with zero Ricci curvature, which is the $\Gamma$-invariant complement of $\mathcal{R}_0(\Gamma)$ in $\mathcal{R}(\Gamma)$; $\pi_{k_1, \dots, k_n}$ denotes the absolutely irreducible representation of the compact Lie algebra $\Gamma$ in a space of dimension $d$, whose complexification is defined by the highest weight $(k_1, \dots, k_n)$; $R_{\mathbb{R}}, R_{\mathbb{C}}, R_{\mathbb{H}}, R_{\mathbb{O}}$ are the curvature tensors of the real, complex, quaternionic projective spaces and the octonionic projective plane, respectively; $J$ is the complex structure operator; $J_\alpha$ ($\alpha=1,2,3$) is a basis of the linear algebra $Sp(1)$ satisfying the conditions $J_\alpha J_\beta = J_\gamma$ (where $(\alpha, \beta, \gamma)$ is a cyclic permutation of indices $(1, 2, 3)$).
Remark. The explicit form of all possible tensors for the remaining irreducible components is omitted due to its complexity.
We now present some corollaries.
Corollary 1. Let $M$ be a Riemannian space with a restricted holonomy group $Spin(9)$. Then $M$ is locally isometric to either the octonionic projective plane $F_4/SO(9)$, its dual non-compact symmetric space, or Euclidean space.
Corollary 2. Let $M$ be an orientable locally irreducible Riemannian space that is not locally symmetric. Suppose that one of the following conditions is satisfied: the Ricci curvature of the space $W^n$ is non-zero; $M$ is compact and admits a one-parameter group of isometries. Then the holonomy group of the space is equal to $U(n)$.
[TABLE:1] Explicit form of some irreducible components of the spaces $\mathcal{R}(\Gamma)$.
$(SO(n)) = \{R_{p,q} : R_{p,q}(X, Y) = X \wedge Y, X, Y \in V\}$
$(U(m)) = \{R_{p,q} = -\frac{1}{4} [J X, J Y] + \frac{1}{4} [X \wedge Y + J X \wedge J Y]\}$
$(Sp(1) \cdot Sp(m)) = \{R_{p,q} : R_{p,q}(X, Y) = -\frac{1}{4} \sum K_i(X, Y) J_i\}$
$(Spin(9)) = \{R_{p,q}\}$
$(SO(n))_1 = \{R_S : R_S(x, y) = Sx \wedge y + x \wedge Sy, S \in V \vee V, \text{sp}(S) = 0\}$
$(U(m))_1 = \{R_S : R_S(x, y) = (Sx, y) J + S J \wedge x - y \wedge SJ - \frac{1}{4} [Sx \wedge y + x \wedge Sy + J Sx \wedge Jy + Jx \wedge J Sy]; S \in V \vee V, \text{sp}(S) = 0\}$
A Riemannian space $V^n$ with holonomy group $\Gamma$ is called Kähler if $\Gamma \subseteq U(m)$, and quaternionic if $\Gamma \subseteq Sp(1) \cdot Sp(m)$. A quaternionic space is said to have constant quaternionic curvature if the curvature $K(x, Ax)$ along two-dimensional sections of the form $(x, Ax)$ is independent of $A \in Sp(1)$ ($|A|=1$) for any tangent vector $x$.
Corollary 3. A quaternionic space with constant quaternionic curvature is locally isometric to either Euclidean space, quaternionic projective space, or quaternionic hyperbolic space.
Corollary 4. Let $V^{4m}$ be a quaternionic space. Then:
1) $V^{4m}$ is an Einstein space. If $V^{4m}$ is compact, irreducible, and admits a one-parameter group of motions, then it has a positively defined Ricci curvature.
2) If the Ricci curvature of the space $V^{4m}$ is non-zero, then the space $V^{4m}$ is irreducible (as a Riemannian space). If, in addition, $V^{4m}$ is not a locally symmetric space, then it has holonomy group $\Gamma = Sp(1) \cdot Sp(m)$.
3) If the Ricci curvature of the space $V^{4m}$ is zero, then $V^{4m}$ is locally isometric to the direct product of a Euclidean space and irreducible quaternionic spaces $V_i$ with holonomy groups $Sp(m_i)$.
Corollary 5. Every locally conformally Euclidean Kähler space is locally Euclidean.
Remark.
1. Some of the statements presented are already known. Name
E. Bonan proved in \cite{3} that a Riemannian manifold with a holonomy group $\Gamma \subset G_2$ or $\Gamma \subset \text{Spin}(7)$ has vanishing Ricci curvature. A similar statement for Riemannian manifolds with holonomy group $\Gamma \subset \text{SU}(m)$ has long been known. These results, combined with Corollary 1, yield Corollary 2. Statement 1 of Corollary 4 was proven by M. Berger \cite{2}, while Statement 3 of Corollary 4 was established by J. Wolf \cite{5}. Statement 2 of Corollary 4
was addressed by D. V. Alekseevsky, who corrected an erroneous proposition contained in the work of J. Wolf \cite{5}. Remark 2: Examples of Riemannian manifolds that are not locally symmetric and have holonomy group $U(m)$ are well known (Kähler manifolds), while those with holonomy group $\Gamma = \text{SU}(m)$ were recently constructed by M. Berger \cite{2}. According to Corollary 1, there are no locally non-symmetric Riemannian manifolds with holonomy group $\Gamma = \text{Spin}(n)$ for $n \neq 7, 8$. The question regarding the existence of locally non-symmetric Riemannian manifolds with holonomy groups $\Gamma = \text{Sp}(1) \cdot \text{Sp}(m)$ ($m > 1$), $\text{Sp}(m)$ ($m > 1$), $G_2$ ($n=7$), or $\text{Spin}(7)$ ($n=8$) remains open.
1. spaces of curvature tensors of type Г
Let $V$ be a Riemannian space with a holonomy algebra $\Gamma$ at a point $p \in V$. As is well known, the curvature tensor of a Riemannian space at a point is a tensor of type (1,3) possessing the following properties:
$$\begin{aligned} 1) & \quad R(x, y) = -R(y, x) \\ 2) & \quad \langle R(x, y)u, v \rangle = -\langle R(x, y)v, u \rangle \\ 3) & \quad R(x, y)z + R(y, z)x + R(z, x)y = 0 \\ 4) & \quad \langle R(x, y)u, v \rangle = \langle R(u, v)x, y \rangle \end{aligned}$$
Here $x, y, z, u, v \in V = V_p$, $\langle \cdot, \cdot \rangle$ is the Euclidean metric of the space $V$, and $R(x, y)$ is an endomorphism of the space $V$ defined by the contraction of the tensor with the vectors $x$ and $y$. We identify the tangent space with its dual space $V^*$ using the Euclidean metric. Then the space of skew-symmetric endomorphisms of the space $V$ is identified with the space $\Lambda^2 V \subset V \otimes V$, and the space of symmetric endomorphisms of the space $\Lambda^2 V$ is identified with the space $S^2(\Lambda^2 V) = \{A \vee B = A \otimes B + B \otimes A \mid A, B \in \Lambda^2 V\}$. Let $R = \sum x_i \wedge y_i \otimes R(x_i, y_i)$. Then conditions 1)–4) show that the tensor can be interpreted as a symmetric endomorphism of the space of skew-symmetric matrices. Moreover, its restriction to the orthogonal complement of the subspace $\Gamma$ in the space $\Lambda^2 V$ is zero. Thus, the tensor can be considered an element of the space $S^2 \Gamma = \{A \vee B = A \otimes B + B \otimes A\}$, the space of symmetric endomorphisms over the Lie algebra $\Gamma$.
The formulas
$$\begin{aligned} b(x \wedge y \vee u \wedge v) &= \frac{1}{3} (R(x, y)u + R(y, u)x + R(u, x)y) \\ r(x \wedge y \vee u \wedge v) &= \frac{1}{2} (\langle x, u \rangle \langle y, v \rangle - \langle x, v \rangle \langle y, u \rangle) \end{aligned}$$
define by linearity two linear maps $b$ and $r$ on the space. A tensor $\xi \in S^2 \Gamma$ is said to satisfy the Bianchi identity if $b(\xi) = 0$. The symmetric tensor $r(\xi)$ is called the Ricci curvature of the tensor $\xi$. For the curvature tensor at a point, the tensor $r(R)$ is the Ricci curvature tensor at that point. Condition 3) means that the curvature tensor satisfies the Bianchi identity.
) That is, those that are not locally symmetric.
*) Author's note: The author has succeeded in constructing an example of a non-symmetric Riemannian space with holonomy group $Sp(1) \cdot Sp(m)$ and classifying all such spaces under the assumption that they admit a simply transitive solvable group of motions.
Let us define $\mathcal{R}(\Gamma) = \{ \xi \in S^2 \Gamma \mid b(\xi) = 0 \}$ and $\mathcal{R}_0(\Gamma) = \{ \xi \in \mathcal{R}(\Gamma) \mid r(\xi) = 0 \}$. The space $\mathcal{R}(\Gamma)$ is called the space of curvature tensors of type $\Gamma$. The following proposition is evident: The curvature tensor $R$ at an arbitrary point $p$ of a Riemannian space with holonomy algebra $\Gamma$ belongs to the space $\mathcal{R}(\Gamma)$. If $R \in \mathcal{R}_0(\Gamma)$, then the Ricci curvature of the space is zero.
We decompose the space $S^2 \Gamma$ into $\Gamma$-irreducible subspaces relative to the action of the Lie algebra $\Gamma$ in $S^2 \Gamma$, which is induced by the adjoint representation. Let us choose a non-zero representative $\xi_i$ from each component $W_i$. Assume that all components are pairwise non-equivalent. Since the maps $b$ and $r$ commute with the action of the Lie algebra $\Gamma$ in the space of tensors over $V$, we have:
$I = \{i : b(\xi_i) = 0\}$ and $I_0 = \{i : b(\xi_i) = 0, r(\xi_i) = 0\}$.
Let $\phi: \Gamma \to \mathfrak{gl}(V)$ be a linear representation of the Lie algebra $\Gamma$ in the space of the smallest dimension. In practice, it is easier to find the decomposition of the space $S^2 \Gamma$ into $\Gamma$-irreducible components $W_i$. To simplify calculations, it is convenient to pass to the complex field, i.e., to seek the spaces $\mathcal{R}(\Gamma^\mathbb{C}) = \{ \xi \in S^2 \Gamma^\mathbb{C} \mid b(\xi) = 0 \}$ and $\mathcal{R}_0(\Gamma^\mathbb{C}) = \{ \xi \in S^2 \Gamma^\mathbb{C} \mid b(\xi) = 0, r(\xi) = 0 \}$, where $b$ and $r$ are the complex extensions of the operators $b$ and $r$, and $\Gamma^\mathbb{C}$ is the complexification of the linear Lie algebra $\Gamma$. Knowing the conjugation of the space $V$ relative to $V^\mathbb{C}$ and the spaces $\mathcal{R}(\Gamma^\mathbb{C}), \mathcal{R}_0(\Gamma^\mathbb{C})$, it is easy to find the spaces $\mathcal{R}(\Gamma)$ and $\mathcal{R}_0(\Gamma)$.
Remark: As shown by M. Berger \cite{1}, the holonomy algebras of pseudo-Riemannian spaces are either complexifications of the holonomy algebras of Riemannian spaces or their real forms. Therefore, the description of the spaces $\mathcal{R}(\Gamma^\mathbb{C})$ for all Lie algebras $\Gamma$ from Berger's list of Riemannian holonomy algebras allows one to easily find all spaces of curvature tensors of a given type for pseudo-Riemannian spaces as well.
§ 2. Description of spaces
Description of the Spaces $J$
We now proceed to describe the spaces $J$. Table 3 presents the canonical bases for the Lie algebras $\Gamma$ and their corresponding bases in the linear Lie algebras $\Gamma^*$ under the isomorphism $\Gamma \to \Gamma^*$ for various simple Lie algebras $\Gamma$ from the Berger list. Additionally, the table includes the symmetric bilinear form of the space $J$, which serves as the complex extension of the Euclidean metric of the space $F$. This bilinear form allows for the identification of the space $J$ with its dual space $J^*$.
[TABLE:3]
The following notation is adopted in the table: $\{e_i\}$ denotes the basis of the space $V$ upon which the Lie algebra $\Gamma$ acts. The indices for these basis elements are defined as follows:
- $i = 1, \dots, m+1$ for $\Gamma = \mathfrak{sl}(m+1, \mathbb{C}) = A_m$;
- $i = \pm 1, \dots, \pm m$ for $\Gamma = \mathfrak{sp}(m, \mathbb{C}) = C_m$;
- $i = 0, \pm 1, \dots, \pm m$ for $\Gamma = \mathfrak{so}(2m+1, \mathbb{C}) = B_m$.
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Riemannian Spaces with Exceptional Holonomy Groups
The indices $i, j$ range from $2, \dots, m$; the indices $\alpha, \beta$ range from $\pm 1, \dots, \pm m$. We define $h_i = A_{i,i} - A_{i+1,i+1}$, $g_{\alpha\beta} = A_{\alpha,-\beta} - A_{\beta,-\alpha}$, and $c_{\alpha\beta} = A_{\alpha,i} A_{i,-\beta} - A_{\alpha,-i} A_{-i,-\beta}$. We denote by $\pi(k_1, \dots, k_r)$ the irreducible representation of a complex simple Lie algebra of rank $r$ in a complex space of dimension $N$, defined by the highest weight $(k_1, \dots, k_r)$. This symbol also denotes the corresponding representation of the compact form of the Lie algebra in a real space of dimension $N$, while $\pi_1$ denotes the one-dimensional trivial representation.
[TABLE:4] provides the decompositions of the spaces $S^2(\Gamma^*) = \Gamma^* \vee \Gamma^*$ into $\Gamma^*$-irreducible components, along with a non-zero representative $\xi$ from each component. If $\Gamma$ is a simple Lie algebra of skew-symmetric transformations of the space $V$, then among the $\Gamma^*$-irreducible components of the space $S^2(\Gamma^*) = \Gamma^* \vee \Gamma^*$, there exists a unique one-dimensional component $\pi_1$ generated by the Killing-Cartan form $g^*$ of the algebra $\Gamma^*$.
A form satisfies the Bianchi identity if and only if there exists a symmetric space with holonomy algebra $\Gamma$. In particular, for the Lie algebras $Sp(m)$ under consideration, the form does not satisfy the Bianchi identity, whereas for the algebras $SO(n)$ it does (the symmetric space $F^4/SO(9)$ has holonomy algebra $Spin(9)$, while the sphere $S^n = SO(n+1)/SO(n)$ has holonomy algebra $SO(n)$). Tables 3 and 4 contain all the necessary information to determine which irreducible components of the space $S^2(\Gamma^*)$ satisfy the Bianchi identity, where $\Gamma$ is a simple Lie algebra from Berger's list.
Let us verify, for example, for which Lie algebras $\Gamma$ the highest component of the decomposition of $S^2(\Gamma^*)$ satisfies the Bianchi identity. The representatives of the highest components take the following forms:
- For $SU(m)$: $\xi_3 = (A_{1,2}) \vee (A_{1,2})$
- For $Sp(m)$: $\xi_3 = (A_{1,-1}) \vee (A_{1,-1})$
- For $G_2$: $\xi_3 = (v_1 \wedge v_2) \vee (v_1 \wedge v_2)$
- For $Spin(7)$: $\xi_3 = (v_1 \wedge v_2 \wedge v_3) \vee (v_1 \wedge v_2 \wedge v_3)$
A tensor of the form $A \vee A$, where $A \in V^* \vee V^*$, satisfies the Bianchi identity if and only if the rank of the matrix $A$ is equal to two. It follows that the highest component of the decomposition of $S^2(\Gamma^*)$ into irreducible $\Gamma$-modules does not satisfy the Bianchi identity for $SU(m)$ and $Sp(m)$, but does satisfy it for $G_2$ and $Spin(7)$. Direct verification shows that all highest components consist of tensors with zero Ricci curvature, and that tensors from other components of dimension greater than 1 do not satisfy the Bianchi identity for Lie algebras $\Gamma \neq SO(n)$. For $\Gamma = SO(n)$, the component $\pi(2,0,\dots,0)$ also satisfies the Bianchi identity.
[TABLE:4]
$V_m \otimes V_m \cong \pi_{(1,0,\dots,0,1)} \oplus \pi_{(0,1,0,\dots,0,1,0)} \oplus \pi_{(2,0,\dots,0,2)}$
$\xi_{1,1} \vee \xi_{1,1} + \xi_{1,2} \vee \xi_{1,n+16}$
$\pi_{(1,0,\dots,0,1)} = A_{1,m+1} \vee A_{1,m} \vee A_{2,m+1}$
$\pi_{(0,1,0,\dots,0,1,0)}$
$\xi_3 = A_{1,2} \vee A_{1,2} \in \pi_{(2,0,\dots,0,2)}$.
$\frac{(n-1)(n+2)}{2} \oplus \frac{n(n-1)(n-2)(n-3)}{(n+2)(n+1)n(n-3)}$
$S^2(B_n) = \pi_1 \oplus \pi_{(0,0,0,1,0,\dots,0)} \oplus \pi_{(0,2,0,\dots,0)}$
$\xi_{1,2} = A_{1,2} \vee A_{3,4} + A_{1,3} \vee A_{2,4} + A_{1,4} \vee A_{2,3} \in \pi_{(0,0,0,1,0,\dots,0)}$,
$\xi_3 = A_{1,2} \vee A_{1,2} \in \pi_{(0,2,0,\dots,0)}$.
$\frac{(n-1)(n+2)}{2} \oplus \frac{n(n-1)(n-2)(n-3)}{(n+2)(n+1)n(n-3)}$
$S^2(D_n) = \pi_1 \oplus \pi_{(0,0,0,1,0,\dots,0)} \oplus \pi_{(0,2,0,\dots,0)}$
$\xi_2 = A_{1,3} \vee A_{1,2} \in \pi_{(1,2,0,\dots,0)}$
$S^2(V_m) \cong \pi_{(0,1,0,\dots,0)} \oplus \pi_{(0,2,0,\dots,0)} \oplus \pi_{(4,0,\dots,0)}$
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§ 2 =
$\mathcal{R}_{1,2} \vee \mathcal{R}_{1,2} = \mathcal{R}_{1,1} \vee \mathcal{R}_{2,2} \oplus \mathcal{R}_{(0,2,0,\dots,0)}$. Let $R(Q) = -\frac{1}{2}(R_1 + R_2) + R_3$, where $R_1 \vee R_{1,3} + R_2 \vee R_{3,1} + R_3 \vee R_{1,2} \in \mathcal{R}_{(0,2)}$. We now turn to the description of the spaces $\mathcal{R}(\mathrm{U}(m))$ and $\mathcal{R}(\mathrm{Sp}(1) \cdot \mathrm{Sp}(m))$. We have:
$$\mathcal{R}(\mathrm{U}(m)) = \mathbb{R} \oplus \mathbb{R} \oplus [\mathcal{R}_1 \vee \mathcal{R}_1] \oplus [\mathcal{R}_1 \vee \mathcal{R}_{\mathrm{SU}(m)(0,1,\dots,0,1)}] \oplus \mathcal{R}_{(0,1,\dots,0,1,0)} \oplus \mathcal{R}_{(2,0,\dots,0,2)}.$$
The two-dimensional subspace $\mathbb{R} \oplus \mathbb{R}$ contains a one-dimensional space satisfying the Bianchi identity, which is generated by the curvature tensor of the complex projective space.
Riemannian spaces with exceptional holonomy groups. Let $e_a, e'_a$ be an orthonormal basis of the space such that $Je_a = e'_a$. The curvature tensor of the complex projective space is given by:
$$R = -2 \sum_{a} (e_a \wedge e'_a) \vee (e_1 \wedge e'_m - e'_1 \wedge e_m)$$
$$-2 (e_1 \wedge e'_1 - e_m \wedge e'_m) \vee (e_1 \wedge e'_1 - e_m \wedge e'_m) + [e_1 \wedge e_a + e'_1 \vee (e_a \wedge e_m + e'_a \wedge e'_m)] - 2 [(e_1 \wedge e'_a - e'_1 \wedge e_a) \vee (e_a \wedge e'_m - e'_a \wedge e_m)]$$
belongs to the space $\mathcal{R}(\mathrm{U}(m)) + \mathcal{R}_{(0,1,\dots,0,1)}$ and satisfies the Bianchi identity. Since the component $\mathcal{R}_{(0,1,\dots,0,1)}$ does not satisfy the Bianchi identity, it follows that the set of tensors in $\mathcal{X}$ satisfying the Bianchi identity forms an irreducible $\mathrm{U}(m)$-module $\mathcal{R}'(\mathrm{U}(m))$ of dimension $m^2-1$, generated by the tensor $R$. Finally, for $\Gamma = \mathrm{Sp}(1) \cdot \mathrm{Sp}(m)$, we have:
$$\mathcal{R} = \mathbb{R} \vee \mathcal{R} = \mathbb{R}_1 \oplus \mathbb{R}_2 + \mathcal{R}_{(0,1,0,\dots,0)} + \mathcal{R}_{(0,2,0,\dots,0)} + \mathcal{R}_0$$
where $\mathrm{Sp}(1) \vee \mathrm{Sp}(1) = 3\mathbb{R}_1 + \mathbb{R}_2$; $\mathbb{R}_1 = \{J_1 \otimes J_1 + J_2 \otimes J_2 + J_3 \otimes J_3\}$ is the trivial $\mathrm{Sp}(1) \cdot \mathrm{Sp}(m)$-module spanned by the Cartan-Killing form of the algebra $\mathfrak{sp}(1)$ with basis $J_1, J_2, J_3$; $\mathbb{R}_2 = \{ \sum a_{ij} J_i \otimes J_j = 0 \}$ is an irreducible $\mathrm{Sp}(1) \cdot \mathrm{Sp}(m)$-submodule, and $\mathbb{R}_0$ is the trivial submodule from $\mathcal{R}(\mathrm{U}(m))$. The two-dimensional $\mathrm{Sp}(1) \cdot \mathrm{Sp}(m)$-module $\mathbb{R}_1 + \mathbb{R}_2$ contains a one-dimensional subspace $\mathcal{R}_P$ satisfying the Bianchi identity. It is generated by the curvature tensor $R_{HP^n}$ of the quaternionic projective space $HP^n = \mathrm{Sp}(m+1)/\mathrm{Sp}(1) \cdot \mathrm{Sp}(m)$. It is evident that the non-trivial $\mathrm{Sp}(1) \cdot \mathrm{Sp}(m)$-submodules of $\mathcal{S}^2(\mathrm{Sp}(1) \cdot \mathrm{Sp}(m))$ are pairwise non-equivalent. By choosing representatives, it is easy to verify that the $\mathrm{Sp}(1) \cdot \mathrm{Sp}(m)$-submodules $\mathbb{R}_1$ and $\mathrm{Sp}(1) \vee \mathrm{Sp}(m)$ satisfy the Bianchi identity if $m=1$ and do not satisfy it if $m > 1$. The results obtained are summarized in Tables 1 and 2.
§ 3. Proof of Corollaries
Proof of the Corollary
Consider the space $\mathcal{R}(Sp(1) + Sp(m))$ and the set of tensors with constant quaternionic curvature: $\mathcal{R}_0 = \{R \in \mathcal{R}(Sp(1) + Sp(m)) : \dots \}$.
The expression $\langle R(x, y)x, J_i x \rangle$ is independent of $J_i \in Sp(1)$. It is evident that $\mathcal{R}_0$ is an $Sp(1) + Sp(m)$-submodule. It is easily verified that tensors from $\mathcal{R}_1(Sp(1) + Sp(m))$ do not belong to the subspace $\mathcal{R}_0$. Consequently, $\mathcal{R}_0 = \mathcal{R}_0(Sp(1) + Sp(m))$; that is, the space is a one-dimensional space spanned by the curvature tensor of the quaternionic projective space $HP^m$. The curvature tensor of a quaternionic space $V^n$ with constant quaternionic curvature at any point belongs to the space $\mathcal{R}_0$ and, therefore, takes the form $\nu(p)R_{HP^m}$. It follows from the second Bianchi identity that $\nu(p)$ is independent of the point $p \in V^n$. The assertion under proof now follows from the Ambrose-Singer theorem, according to which a complete, simply connected Riemannian manifold is determined (up to isometry) by its curvature tensor field.
D. V. Alekseevsky
Corollaries 1, 2, and point 1 of Corollary 4 are proved in a similar manner.
Proof of points 2) and 3) of Corollary 4.
Point 2). Let the Ricci curvature of the quaternionic space $V^{4m}$ be non-zero. Then, according to Tables 1 and 2, the curvature tensor of the space at an arbitrary point has the form $R(x, y) = \nu \sum \langle x, J_\alpha y \rangle J_\alpha + P(x, y)$, where $P(x, y) \in Sp(m)$ and $x, y \in V_p$. Therefore, the holonomy algebra $\Gamma$ of the space contains transformations of the form $\sum \lambda_\alpha J_\alpha + A$, where $A \in Sp(m)$ and $\{J_\alpha\}$ is a basis of $Sp(1)$. Consequently, it decomposes into a sum of two ideals $\Gamma = M + N$, where $N \subset Sp(m)$ and $J_\alpha + Z_\alpha \in M$.
Suppose that the space is reducible. Since the Ricci tensor is non-degenerate, according to de Rham's theorem, the holonomy algebra $\Gamma$ can be decomposed into a sum of two non-zero ideals $\Gamma = \Gamma_1 \oplus \Gamma_2$, and the space $V_p$ into a direct sum of two non-trivial subspaces $V_1$ and $V_2$ such that $\Gamma_i V_j = 0$ for $i \neq j$. Since $[Sp(1), \Gamma_2] = 0$, we have $Sp(1)V_1 \subset V_1$ and $Sp(1)V_2 \subset V_2$. We thus find that the $Sp(1)$-invariant subspace $V_1$ is annihilated by the transformations $J_\alpha + Z_\alpha$, which is impossible. This contradiction proves the first part of assertion 2). The second part follows from Berger's theorem.
Point 3). To prove point 3), it is sufficient to verify that all irreducible components of the de Rham decomposition of a quaternionic space $V$ with zero Ricci curvature are themselves quaternionic spaces. Let $V = V_0 + V_1 + \dots + V_s$ and $\Gamma = \Gamma_1 + \dots + \Gamma_s$ be the decompositions of the space and the holonomy algebra corresponding to the de Rham decomposition, where $\Gamma_i V_j = 0$ for $i \neq j$ and $\Gamma_0 V_0 = 0$. Since $[Sp(1), \Gamma] = 0$, each $V_i$ is an $Sp(1)$-invariant subspace and $\Gamma_i \subset Sp(1) + Sp(m_i)$, which proves assertion 3).
Proof of the Corollary
5. Let
Let $R \in \mathcal{R}(SO(n))$ be the Riemann curvature tensor of a space $V^n$ at an arbitrary point. The projection of this tensor onto the subspace $\mathcal{W}(SO(n))$ parallel to the subspace $\mathcal{R}'(SO(n)) + \mathcal{R}''(SO(n))$ is called the Weyl tensor of the space $V^n$ at that point. According to the classical result of H. Weyl, the space $V^n$ ($n > 3$) is locally conformally Euclidean if and only if its Weyl tensor vanishes at all points. This occurs when, at any given point, the curvature tensor takes the form $R(x, y) = Sx \wedge y + x \wedge Sy$, where $S$ is a symmetric matrix. However, as is easily seen, a non-zero tensor of this form cannot belong to the space $\mathcal{R}(U(m))$. This completes the proof of the assertion.
Scientific Research Institute of Organic Intermediates and Dyes
Received by the Editorial Board on April 29, 1967.
REFERENCES
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C. R. Acad. Sci. 262 (1966), 1316–1318. Bonan, E., "Sur les variétés riemanniennes à groupe d'holonomie $G_2$ ou $Spin(7)$," C. R. Acad. Sci. 262 (1966), 127–129. J. Math. Phys. 2, No. 3 (1961), 327–332.
Wolf J. A., Complex homogeneous contact manifolds and quaternionic symmetric spaces, J. Math. Mech. 14 (1965), 1033—1047.