I am studying the evolution of curvature in my study on the Ricci flow, and in The Ricci Flow in Riemannian Geometry by Hopper and Andrews, I came across the (0,4) quadratic curvature tensor defined by $$B(X,Y,W,Z) = \langle R(X,\cdot,Y,\star \rangle,R(W,\cdot,Z,\star ) \rangle,$$ which in components becomes \begin{equation}B_{ijkl} = g^{pr}g^{qs}R_{piqj}R_{rksl}. \end{equation} Here $R$ denotes the Riemann curvature tensor. They use that the inner product for two smooth tensor fields $\alpha, \beta \in \mathcal{T}_s^r(M)$ on a smooth manifold $M$ is given by $$\langle \alpha, \beta \rangle = g^{a_1 b_1}\cdots g^{a_r b_r} g_{i_1j_1} \cdots g_{i_s j_s} \alpha_{a_1 \cdots a_r}^{i_1 \cdots i_s} \beta_{b_1 \cdots b_r}^{j_1 \cdots j_s}.$$

I understand this definition, but when I'd apply it to the inner product of the (0,4) Riemann curvature tensor $R$, for the components I would get that \begin{align} B_{ijkl} & = \langle R_{ipjq}dx^i\otimes dx^p \otimes dx^j \otimes dx^q, R_{krls}dx^k\otimes dx^r \otimes dx^l \otimes dx^s \rangle \\ & = g^{ik}g^{pr}g^{jl}g^{qs}R_{ipjq}R_{krls}. \end{align}

So, I'm missing a point somewhere. I doubt whether I understand the notation correctly with the $\cdot,\star$. Are these meant to be left out in such a derivation? Or what else would be going wrong here?

  • $\begingroup$ shouldn't your last factor of $g$ in your result read $g^{qs}$ not $g^{rs}$? $\endgroup$
    – user284001
    May 18, 2021 at 10:20
  • $\begingroup$ Yes indeed, thanks for noting. $\endgroup$
    – Luukdr
    May 18, 2021 at 10:21

1 Answer 1


First, note that $R(X, \cdot, Z, \star)$ is the $(2, 0)$ (or $(0, 2)$ or $(1, 1)$, doesn't matter in the end) tensor field which takes as input $2$ vector fields $(Y, W)$ and returns as its output the real valued function $R(X, Y, Z, W)$, which in turn takes as inputs a point $p$ and spits out $R(p)(X_p, Y_p, Z_p, W_p) \in \mathbb{R}$. Equivalently, in local coordinates it's the tensor given by

$$\alpha_{k \ell} = R_{ikj \ell} \mathrm{d} x^{i} \otimes \mathrm{d} x^j$$ This is all very formal and all, but you mentioned you were insecure with it so I thought I'd explain. Now, we want to calculate $\langle \alpha, \beta \rangle$, where $\alpha = R(\partial_i, \cdot, \partial_j, \star)$ and $\beta =R(\partial_k, \cdot, \partial_{\ell}, \star) $. By definition, we have:

$$\begin{aligned} B_{ijk \ell} &= \langle R(\partial_i, \cdot, \partial_j, \star),R(\partial_k, \cdot, \partial_{\ell}, \star) \rangle \\ &= g^{a_1 b_1} g^{a_2 b_2} \alpha(a_1, a_2) \beta(b_1, b_2) \\ &= g^{pr} g^{qs} \alpha(p, q) \beta(r, s) \\ &= g^{pr} g^{qs} R_{ipjq} R_{k r \ell s} \\ &= g^{pr} g^{qs} R_{piqj} R_{rk s \ell }\end{aligned} $$

where I just switched $a_1, b_1, a_2, b_2$ for $p, r, q, s$ and used the skew-symmetry of the curvature tensor in the last step. Your mistake is that you wrote $\alpha = R_{ipjq} \mathrm{d} x^i\otimes \mathrm{d}x^p \otimes \mathrm{d}x^j \otimes \mathrm{d}x^q$: if this were true, $\alpha $ would be a $(4, 0)$ tensor, which is false - as I mentioned, it's actually a $(2, 0)$ tensor (and you made the same mistake with $\beta$).

If you're still a little insecure with all this, I highly reccomend Ivo Terek's text on tensors (he explains it all very well - actually all his texts are really good, it's worth it taking the time to read the other ones he puts on his personal webpage).


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