Motivation. Let $M$ be a Riemannian submanifold of $(Q, \langle \cdot{,} \cdot \rangle)$, and let $\alpha$ be the (vector-valued) second fundamental form of $M$, i.e., the map $TM \times TM \to NM$ defined by
$$
(X, Y) \mapsto \pi^{\perp}\nabla_{X}Y,
$$
where $\pi^{\perp}$ denotes orthogonal projection onto the normal bundle and $\nabla$ is the ambient Levi-Civita connection.
Let $p \in M$. If we choose a local frame $N_{1}, \dotsc, N_{k}$ of unit normal vector fields around $p$, then we can define $k$ scalar second fundamental forms $h_{1}, \dotsc, h_{k}$ by
$$
h_{i} = \langle N_{i}, \alpha \rangle.
$$
Then $h_{i}$ defines a quadratic form $TM \to \mathbb{R}$, which takes $X$ to
$$
h_{i}(X, X) = \langle N_{i}, \alpha(X,X) \rangle = \langle N_{i}, \pi^{\perp}\nabla_{X}X \rangle = \langle N_{i}, \nabla_{X}X \rangle = -\langle \nabla_{X} N_{i}, X \rangle.
$$
I wonder if also $X \mapsto \lVert \alpha(X, X) \rVert$ defines a quadratic form on $M$.
Question. Let $V$ be a vector space, let $W$ be an inner product space, and let $\alpha$ be a symmetric bilinear form $V\times V \to W$. Is the map
$$\begin{aligned}
\beta \colon V &\to \mathbb{R}\\
v &\mapsto \lVert \alpha(v,v) \rVert
\end{aligned}$$
a quadratic form?
Some comments. A quadratic form on $V$ is a map $q \colon V \to \mathbb{R}$ such that $q(\lambda v) = \lambda^{2}q(v)$ for all $(\lambda,v) \in \mathbb{R}\times V$ and the function $q(u+v) – q(u) – q(v)$ is bilinear.
It is easy to see that $\beta$ satisfies the first condition. However, I am not able to say much about the second one, i.e., to either prove or disprove that $b(u,v) = \beta(u+v) – \beta(u) -\beta(v)$ is bilinear.
For instance, computing $b(-u, v)$, one gets
$$
2b(-u,v) = \lVert \alpha(u, u) + \alpha(v,v) – 2\alpha(u,v) \rVert – \lVert \alpha(u,u) \rVert – \lVert \alpha(v,v) \rVert,
$$
whereas
$$
-2b(u,v) = -\lVert \alpha(u, u) + \alpha(v,v) + 2\alpha(u,v) \rVert + \lVert \alpha(u,u) \rVert + \lVert \alpha(v,v) \rVert.
$$
Best Answer
$$V=\mathbb R^2,\quad W=\mathbb R^1$$ $$\alpha(u,v)=u_1v_1-u_2v_2$$ $$\beta(v)=|v_1^2-v_2^2|$$
Any quadratic form is differentiable, but this is not differentiable.
Here's a slightly more interesting counter-example.
$$V=W=\mathbb R^2$$ $$\alpha(u,v)=(u_1v_1,u_2v_2)$$ $$\beta(v)=\lVert(v_1^2,v_2^2)\rVert=\sqrt{v_1^4+v_2^4}$$ $$\beta(v)\overset?=c_1v_1^2+c_2v_1v_2+c_3v_2^2$$ $$c_1=\beta((1,0))=\sqrt{1^4+0^4}=c_3=\beta((0,1))=\sqrt{0^4+1^4}=1$$ $$c_1+c_2+c_3=\beta((1,1))=\sqrt{1^4+1^4}=\sqrt2$$ $$\beta(v)=1\,v_1^2+(\sqrt2-2)v_1v_2+1\,v_2^2$$ $$\beta((2,1))=2^2+(\sqrt2-2)2+1=2\sqrt2+1$$ $$=\sqrt{2^4+1^4}=\sqrt{17}$$