What are the eigenvalues of the curvature operator on $\mathbb{C}P(2)$?

Question

I tried asking this on stackexchange but was unsuccessful.

On page 150 of section 4.5.3 of Peter Petersen's Riemannian Geometry it is noted that, given an orthonormal basis $X,iX,Y,iY$ for $T_p\mathbb{C}P^2$, the following basis diagonalizes the curvature operator $\mathfrak{R}:\Lambda^2T_p\mathbb{C}P^2 \to \Lambda^2T_p\mathbb{C}P^2 $:

\begin{align*} &X \wedge iX \pm Y \wedge iY\\ & X \wedge Y \pm iX \wedge iY \\ & X \wedge iY \pm Y \wedge iX \end{align*}

with eigenvalues lying in $[0,6]$. I have attempted the calculations using the O'Neill formula but get stuck. For example assume $\mathfrak{R}(X \wedge iX + Y \wedge iY)=c \left(X \wedge iX + Y \wedge iY\right)$. Then:

\begin{align*} g(\mathfrak{R}(X \wedge iX + Y \wedge iY),X \wedge iX + Y \wedge iY)&=cg(X \wedge iX + Y \wedge iY,X \wedge iX + Y \wedge iY)\\ &=2c \end{align*}

and by the definition of $\mathfrak{R}$:

\begin{align*} &g(\mathfrak{R}(X \wedge iX + Y \wedge iY),X \wedge iX + Y \wedge iY) \\ &=R(X,iX,iX,X)+R(Y,iY,iY,Y)+2R(X,iX,iY,Y) \\ &=\sec(X,iX)+\sec(Y,iY)+2R(X,iX,iY,Y) \\ &=8+2R(X,iX,iY,Y) \end{align*}

so:

\begin{align*}2c &= 8+2R(X,iX,iY,Y) \\ c&=4+R(X,iX,iY,Y) \end{align*}

At this point I do not know how to proceed. I know of a formula for expanding $R(X,iX,iY,Y)$ in terms of sectional curvatures but it is quite complicated. Alternatively, going back to the O'Neill formula we have:

\begin{align*}R(X,iX,iY,Y)=&\overline{R}(\overline{X},\overline{iX},\overline{iY},\overline{Y})+\frac14 \overline{g} ([\overline{iX},\overline{Y} ],[\overline{X},\overline{iY}])-\frac14 \overline{g}([\overline{X},\overline{Y}],[\overline{iX},\overline{iY}]) \\ &+ \frac12 \overline{g} ([ \overline{X},\overline{iX}],[\overline{iY},\overline{Y}] )\end{align*}

where $\overline{R}$ denotes the curvature tensor on $S^5$, $\overline{g}$ denotes the metric on $S^5$, and $\overline{V}$ denotes a horizontal lift. Since $X,iX,Y,iY$ are orthonormal and their lifts are also orthonormal I suppose $\overline{R}(\overline{X},\overline{iX},\overline{iY},\overline{Y})=0$ so we are left with

\begin{align*}R(X,iX,iY,Y)=&\frac14 \overline{g} ([\overline{iX},\overline{Y} ],[\overline{X},\overline{iY}])-\frac14 \overline{g}([\overline{X},\overline{Y}],[\overline{iX},\overline{iY}]) + \frac12 \overline{g} ([ \overline{X},\overline{iX}],[\overline{iY},\overline{Y}] )\end{align*}

Have I missed something that should make this easier to work out?

edit: I followed through with the first idea (expanding out $R(X,iX,iY,Y)$ in terms of sectional curvatures) and got that the eigenvalues are $0,0,2,2,2,6$. Given that $\mathbb{C}P(2)$ is Einstein with Einstein constant $6$ and Scalar curvature $12$, I am tempted to believe that the eigenvalues are correct (as their sum does equal $12$).

Answer 1

Perhaps, it would be easier to just compute it directly from the structure equations. For example, suppose we wanted to compute the eigenvalues of the curvature operator for $\mathbb{CP}^n=\mathrm{SU}(n{+}1)/\mathrm{U}(n)$. I claim that they are $0$ with multiplicity $n(n{-}1)$, $2$ with multiplicity $n^2{-}1$, and $2(n{+}1)$ with multiplicity $1$. Here's how to see this: Write the left-invariant form on $\mathrm{SU}(n+1)$ as $$ \lambda = \begin{pmatrix} -i\,n\,\rho & -^t\omega+i\,^t\eta\\ \omega+i\,\eta &\alpha + i\rho\,I_n + i\,\beta\end{pmatrix} $$ where $\omega$ and $\eta$ are columns of height $n$, $\alpha = -\ ^t\alpha$ and $\beta = {}^t\beta$ while $\mathrm{tr}(\beta)=0$. The pullback of the metric on $\mathbb{CP}^n$ to $\mathrm{SU}(n{+}1)$ is the quadratic form $g = \ ^t\omega\circ\omega + ^t\eta\circ\eta$. The Maurer-Cartan equation $\mathrm{d}\lambda = -\lambda\wedge\lambda$ unpacks to $$ \mathrm{d}\begin{pmatrix}\omega\\ \eta\end{pmatrix} =-\begin{pmatrix}\alpha & -(n{+}1)\rho I_n-\beta \\ (n{+}1)\rho I_n+\beta & \alpha \end{pmatrix}\wedge \begin{pmatrix}\omega \\ \eta\end{pmatrix} $$ together with the equations $$ \begin{align} \mathrm{d}\rho &= -2/n\,{}^t\omega\wedge\eta\\ \mathrm{d}\alpha + \alpha\wedge\alpha - \beta\wedge\beta &= \omega\wedge {}^t\omega + \eta \wedge {}^t\eta\\ \mathrm{d}\beta +\beta\wedge\alpha-\alpha\wedge\beta &= \eta\wedge {}^t\omega- \omega\wedge {}^t\eta + (2/n) {}^t\omega\wedge\eta\, I_n\,. \end{align} $$ Consequently, the matrix $$ \theta = \begin{pmatrix}\alpha & -(n{+}1)\rho I_n-\beta \\ (n{+}1)\rho I_n+\beta & \alpha \end{pmatrix} $$ is the Levi-Civita connection matrix for the metric $g$, and its curvature is $$ \begin{align} \Theta &= \mathrm{d}\theta+\theta\wedge\theta\\ &= \begin{pmatrix}\omega\wedge {}^t\omega + \eta \wedge {}^t\eta & 2\,{}^t\omega\wedge\eta\,I_n -\eta\wedge {}^t\omega+\omega\wedge {}^t\eta \\ -2\,{}^t\omega\wedge\eta\,I_n +\eta\wedge {}^t\omega- \omega\wedge {}^t\eta & \omega\wedge {}^t\omega + \eta \wedge {}^t\eta \end{pmatrix}. \end{align} $$ Now, $\theta$ (and hence $\Theta$) takes values in the subalgebra ${\frak{u}}(n)\subset{\frak{so}}(2n)\simeq\Lambda^2(\mathbb{R}^{2n})$, where ${\frak{u}}(n) = {\mathbb{R}}\cdot J + {\frak{su}}(n)$ with $J = \begin{pmatrix} 0_n & I_n\\ -I_n & 0_n\end{pmatrix}$ and ${\frak{su}}(n)$ being the skew-symmetric matrices of the form $\begin{pmatrix} a & b\\ -b & a\end{pmatrix}$ where $a$ is $n$-by-$n$ skew-symmetric and $b$ is $n$-by-$n$ symmetric and tracefree. It follows that the curvature operator $R$ (which is symmetric) annihilates everything in ${\frak{u}}(n)^\perp\subset {\frak{so}}(2n)$, a vector space of dimension $n(n{-}1)$, so these kernel eigenvectors all have eigenvalue $0$. Taking the $J$-trace of $\Theta$, we see that $R$ has $J$ as an eigenvector of eigenvalue $2(n{+}1)$. The subspace ${\frak{su}}(n)$ is irreducible and must be mapped by $R$ to a scalar multiple of itself, and pairing with a typical element, we see that the multiplier is $2$. Thus, $2$ is an eigenvalue of multiplicity $n^2{-}1$, the dimension of ${\frak{su}}(n)$.