The sectional curvature of this explicitely given n-dimensional submanifold of R^(n+1)

curvaturedifferential-geometry

Let $n>1$ be an integer. Consider $\mathbb{R}^{n+1}$ with the coordinates $x_1,\ldots,x_n,y$ and the closed submanifold
$$
M_n := \left\{ (x_1,\ldots,x_n,y)\in\mathbb{R}^{n+1}\ \mid\ y \sum_{j=1}^n x_j^2 = 1\right\}.
$$
Since it is a submanifold of Euclidean space with its canonical flat metric, $M_n$ becomes a Riemannian manifold. I want to compute its sectional curvature (for a given point $(x,y)\in M_n$ and a two vectors $v,w\in T_{(x,y)}M_n$), but do not know how to proceed. For $n=2$, sectional curvature is the same as usual Gaussian curvature and I found a formula for computing it, but for higher dimensions my differential geometry skills are not enough.
It is clear that $M_n$ is the graph of a smooth function on $\mathbb{R}^n\setminus\{0\}$ and it is "rotational invariant", in the sense that the value of the curvature at a point $(x,y)$ should only depend on $\|x\|=\sqrt{\sum_{j=1}^n x_j^2}$, but I don't know if that is of any help.

What I would expect to happen is that the sectional curvature is everywhere non-positive.
But I would like to know wether the sectional curvature of the manifold is bounded below (I would guess yes, because that is what happend for $n=2$) and whether there is a lower bound that does not depend on $n$ (I have no idea what to guess here).

Any help would be greatly appreciated.

Best Answer

$\newcommand{\Two}{\mathsf{I}\mathsf{I}}$ $\newcommand{\rD}{\mathrm{D}}$ You can use the second fundamental form and the Gauss-Codazzi equation. For embedded manifolds in a vector space, the Gauss-Codazzi equation reads: $$R_{vw}w.v = \Two(v, v). \Two(w, w) - \Two(v,w). \Two(v,w) $$ and the sectional curvature is $$sec(v, w) = \frac{R_{vw}w.v}{(v.v)(w.w) - (v.w)^2}. $$ The formula for the second fundamental form with the embedded metric is simply $$\Two(v, w) = (\rD_v\Pi)w = (\rD_w\Pi)v$$ where $\Pi$ is the metric compatible projection to the tangent space. The second fundamental form is a covariant derivative of the projection, but it just works out to that directional derivative if you use the embedded metric in a vector space.

Let $z = (x, y)$ satisfying the defining equation and $C(z) = yx^Tx-1$, then its Jacobian is $$J:\omega=(\omega_x, \omega_y) \mapsto 2y(x^T\omega_x)+ \omega_yx^Tx $$ or in matrix form $J$ is the $1\times (n+1)$ matrix $(2yx^T, x^Tx)$ $$\Pi(z)\omega = \omega - J^T(JJ^T)^{-1}J\omega.$$ Here, $JJ^T = |J|^2 = 4y^2x^Tx + (x^Tx)^2$ $$\Pi(z)\omega = \begin{bmatrix}\omega_x\\ \omega_y\end{bmatrix} - \frac{2y(x^T\omega_x)+ \omega_yx^Tx}{4y^2x^Tx + (x^Tx)^2}\begin{bmatrix}2yx\\ x^Tx\end{bmatrix} $$ if $v=(v_x, v_y), w = (w_x, w_y)$ are two tangent vectors then $$\Two(v, w) =- \frac{2v_y(x^Tw_x)+ 2y(v_x^Tw_x) + 2 w_yx^Tv_x}{4y^2x^Tx + (x^Tx)^2}\begin{bmatrix}2yx\\ x^Tx\end{bmatrix} $$ noting that since $w$ is a tangent vector, when you take directional derivative, any term that leaves the Jacobian intact will evaluate to zero. From here, you can evaluate the sectional curvature. You can simplify using $v_y =-\frac{2yx^Tv_x}{x^Tx}$.

See https://arxiv.org/abs/2307.10017 for some other examples of this type of calculation.

As this type of question appears quite often, let's give a general answer. If a hypersurface $M$ is given by a level set equation $\phi(x) = C$ in an inner product space $E$ with $\phi$ is a scalar function, let $g_{\phi}$ be its gradient and $hess_{\phi}$ be its hessian, consider as a bilinear form, so $hess_{\phi}(\xi, \eta)$ is a number for two vectors $\xi, \eta$. Then under the embedded metric

  1. The Levi-Civita connection is $$\nabla_XY = D_XY + \frac{hess_{\phi}(X, Y)}{|g_{\phi}|^2}g_{\phi}$$ Here, two vector fields $X, Y$ on $M$ are considered as vectors in $E$, and $D_XY$ is the flat connection on $E$, ie simple Lie derivative.
  2. The second fundamental form is $-\frac{hess_{\phi}(X, Y)}{|g_{\phi}|^2}g_{\phi}$.
  3. The sectional curvature numerator is $$R_{XY}Y.X = \frac{1}{|g_{\phi}|^2} \{hess_{\phi}(X, X)hess_{\phi}(Y,Y)- hess_{\phi}(X, Y)^2\}. $$

Proof: For 1, the right hand side is a vector field, as differentiating $g_{\phi}^ . Y = 0$ using $X$, we get $$hess_{\phi}(X, Y) + g_{\phi}. D_XY = 0$$ which implies $g_{\phi}. \{D_XY + \frac{hess_{\phi}(X, Y)}{|g_{\phi}|^2}g_{\phi}\} =0$, it satisfies the usual conditions of a covariant derivative, and metric compatibility follows from $$2Y . (D_XY + \frac{hess_{\phi}(X, Y)}{|g_{\phi}|^2}g_{\phi}) = 2Y . D_XY= D_X (Y. Y) $$ as $Y.g_{\phi} = 0$.

For 2, it is well-known $\nabla_XY = \Pi D_XY$ if $\Pi$ is the metric-compatible projection, hence $\Two(X, Y) = D_XY - \Pi D_XY = D_XY - \nabla_XY=-\frac{hess_{\phi}(X, Y)}{|g_{\phi}|^2}g_{\phi} $. The Gauss-Codazzi equation gives 3.

Related Question