# Mean curvature vector

On a Riemannain manifold $$M$$ with a submanifold $$S$$, we can define the second fundamental form $$H$$ to be $$-H(X, Y) = (\nabla_{X} Y)^{N}$$ the component of $$(\nabla_{X} Y)^{N}$$ normal to $$S$$. Then mean curvature vector is defined to be the trace of the second fundamental form. I am confused as to what does it mean for a vector to be the trace of the second fundamental form.

Usually contractions/trace will yield a real-valued function on the manifold, but here we want a vector.

• The mean curvature is the trace. The mean curvature vector is the trace $\times$ the normal vector field. Feb 19, 2021 at 9:57
• The trace of a 2-tensor is a scalar, but the second fundamental form of a submanifold (possibly with codimension > 1) is not a 2-tensor. It is a vector-valued 2-tensor. So its trace is a vector. Feb 19, 2021 at 15:38
• How do you take the trace of a vector valued tensor? Are there any references?@Deane
– z.z
Feb 19, 2021 at 17:02
• The definition of the trace is essentially the same. You simply define the trace of $H$ to be $$\operatorname{tr} H = \sum_{k=1}^n H(e_k,e_k)$$ where $(e_1, \dots, e_n)$ is an orthonormal basis of $T_pM$. Or, if you have coordinate vector fields $\partial_1, \dots, \partial_n$ and dual $1$-forms $dx^1, \dots, dx^n$ on $M$ and the metric is $g_{ij}dx^idx^j$, then $$\operatorname{tr} H = g^{ij}H(\partial_i,\partial_j)$$ If a basis of the normal bundle is $\nu_1, \dots, \nu_{m-n}$, then you can write $H = H^\mu_{ij}dx^idx^j\nu_\mu$ and its trace as $g^{ij}H_{ij}^\mu\nu_\mu$. Feb 19, 2021 at 18:02
• Thanks for the clarification. Do you want to make that into an answer so that I can accept it?
– z.z
Feb 19, 2021 at 21:50

The definition of the trace is essentially the same. You simply define the trace of 𝐻 to be $$\operatorname{tr} 𝐻= \sum_{k=1}^n H(e_k,e_k)$$ where $$(𝑒_1,…,𝑒_𝑛)$$ is an orthonormal basis of $$𝑇_𝑝𝑀$$. Or, if you have coordinate vector fields $$\partial_1,\dots,\partial_𝑛$$ and dual 1-forms $$𝑑𝑥^1,\dots,𝑑𝑥^𝑛$$ on 𝑀 and the metric is $$g_{ij}dx^i\,dx^j$$, then $$\operatorname{tr} 𝐻= g^{ij} H(\partial_i,\partial_j)$$ If a basis of the normal bundle is $$\nu_1,\dots,\nu_{n-m}$$, then you can write $$𝐻=\nu_\mu H^\mu_{ij}dx^idx^j$$ and its trace as $$𝑔^{ij}𝐻^\mu_{ij}\nu_\mu$$
• The scalar mean curvature is defined only if the codimension of $S$ is 1. In that case, the unit normal $\nu$ (also, known as the Gauss map) is unique up to sign. The scalar mean curvature is $\nu\cdot \operatorname{tr} H$. Feb 23, 2021 at 5:49