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Carlo Beenakker
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I'm trying understand a computation on page 371 of Besse's book on (p. 371)Einstein Manifolds.

I already know the curvature operator $R:\bigwedge^2\to\bigwedge^2$ may be written in block diagonal form relative to the direct sum decomposition: $$R=\begin{pmatrix}A&B\\C&D\end{pmatrix}$$ where $A=A^*,C=B^*,D=D^*$. And, $$\begin{pmatrix}A&0\\0&D\end{pmatrix}-s/12=W,\mbox{the Weyl tensor}$$ The two components of the Weyl tensor $W^+=A-s/12,W^-=D-s/12$ are called the self-dual and the anti-self-dual parts respectively.

Now, pay attetion in this part:

$$p_1(M)=-\frac{1}{8\pi^2}\int_MTr(R\wedge R)$$ where $R$ is considered as a matrix of 2-forms. Since $B$ and $B^*$ are acting on orthogonal spaces(this part is OK), $$\begin{matrix}Tr(R\wedge R)&=&Tr(A\wedge A)+Tr(D\wedge D)\\&=&-2(|W^+|^2-|W^-|^2)\omega_g\end{matrix}$$ because $\alpha\wedge\alpha=|\alpha|^2\omega_g$ if $\alpha$ is sel-dual.

My questions are:

  1. What $Tr(R\wedge R)$ stand for?

  2. How compute $Tr(R\wedge R)$, and why the term -2 appears in the expression?

I greatly appreciate any response, observation or correction!

I'm trying understand a computation on Besse's book (p. 371).

I already know the curvature operator $R:\bigwedge^2\to\bigwedge^2$ may be written in block diagonal form relative to the direct sum decomposition: $$R=\begin{pmatrix}A&B\\C&D\end{pmatrix}$$ where $A=A^*,C=B^*,D=D^*$. And, $$\begin{pmatrix}A&0\\0&D\end{pmatrix}-s/12=W,\mbox{the Weyl tensor}$$ The two components of the Weyl tensor $W^+=A-s/12,W^-=D-s/12$ are called the self-dual and the anti-self-dual parts respectively.

Now, pay attetion in this part:

$$p_1(M)=-\frac{1}{8\pi^2}\int_MTr(R\wedge R)$$ where $R$ is considered as a matrix of 2-forms. Since $B$ and $B^*$ are acting on orthogonal spaces(this part is OK), $$\begin{matrix}Tr(R\wedge R)&=&Tr(A\wedge A)+Tr(D\wedge D)\\&=&-2(|W^+|^2-|W^-|^2)\omega_g\end{matrix}$$ because $\alpha\wedge\alpha=|\alpha|^2\omega_g$ if $\alpha$ is sel-dual.

My questions are:

  1. What $Tr(R\wedge R)$ stand for?

  2. How compute $Tr(R\wedge R)$, and why the term -2 appears in the expression?

I greatly appreciate any response, observation or correction!

I'm trying understand a computation on page 371 of Besse's book on Einstein Manifolds.

I already know the curvature operator $R:\bigwedge^2\to\bigwedge^2$ may be written in block diagonal form relative to the direct sum decomposition: $$R=\begin{pmatrix}A&B\\C&D\end{pmatrix}$$ where $A=A^*,C=B^*,D=D^*$. And, $$\begin{pmatrix}A&0\\0&D\end{pmatrix}-s/12=W,\mbox{the Weyl tensor}$$ The two components of the Weyl tensor $W^+=A-s/12,W^-=D-s/12$ are called the self-dual and the anti-self-dual parts respectively.

Now, pay attetion in this part:

$$p_1(M)=-\frac{1}{8\pi^2}\int_MTr(R\wedge R)$$ where $R$ is considered as a matrix of 2-forms. Since $B$ and $B^*$ are acting on orthogonal spaces(this part is OK), $$\begin{matrix}Tr(R\wedge R)&=&Tr(A\wedge A)+Tr(D\wedge D)\\&=&-2(|W^+|^2-|W^-|^2)\omega_g\end{matrix}$$ because $\alpha\wedge\alpha=|\alpha|^2\omega_g$ if $\alpha$ is sel-dual.

My questions are:

  1. What $Tr(R\wedge R)$ stand for?

  2. How compute $Tr(R\wedge R)$, and why the term -2 appears in the expression?

I greatly appreciate any response, observation or correction!

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Geom math
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The trace of a wedge product of matrices

I'm trying understand a computation on Besse's book (p. 371).

I already know the curvature operator $R:\bigwedge^2\to\bigwedge^2$ may be written in block diagonal form relative to the direct sum decomposition: $$R=\begin{pmatrix}A&B\\C&D\end{pmatrix}$$ where $A=A^*,C=B^*,D=D^*$. And, $$\begin{pmatrix}A&0\\0&D\end{pmatrix}-s/12=W,\mbox{the Weyl tensor}$$ The two components of the Weyl tensor $W^+=A-s/12,W^-=D-s/12$ are called the self-dual and the anti-self-dual parts respectively.

Now, pay attetion in this part:

$$p_1(M)=-\frac{1}{8\pi^2}\int_MTr(R\wedge R)$$ where $R$ is considered as a matrix of 2-forms. Since $B$ and $B^*$ are acting on orthogonal spaces(this part is OK), $$\begin{matrix}Tr(R\wedge R)&=&Tr(A\wedge A)+Tr(D\wedge D)\\&=&-2(|W^+|^2-|W^-|^2)\omega_g\end{matrix}$$ because $\alpha\wedge\alpha=|\alpha|^2\omega_g$ if $\alpha$ is sel-dual.

My questions are:

  1. What $Tr(R\wedge R)$ stand for?

  2. How compute $Tr(R\wedge R)$, and why the term -2 appears in the expression?

I greatly appreciate any response, observation or correction!