# Question regarding the quadratic curvature tensor

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 $$$$B_{ijkl} = g^{pr}g^{qs}R_{piqj}R_{rksl}.$$$$ Here $$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?

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

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 $$\beta$$).