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Riemannian Geometry is a subfield of Differential Geometry, which specifically studies "Riemannian Manifolds", manifolds with "Riemannian Metrics", which means that they are equipped with continuous inner products.
2
votes
Riemann's formula for the metric in a normal neighborhood
There is the following way of doing it. Let $\Gamma_{ijk}\equiv g_{is}\Gamma^{s}_{jk}$. Then (see here)
$$
g_{jk}(x)=\delta_{jk}+\int_0^1 \alpha ~x^i\Gamma_{[ji]k}(\alpha x)~d\alpha
$$
If we expand $\ …
3
votes
0
answers
203
views
Volume of sub-manifold of $\mathbf R^n$
Let $M$ be an $n-m$ dimensional sub-manifold of $\mathbf R^n$ defined by the following set of polynomial equations:
\begin{equation}
P_1(\vec x)=0, \\
\vdots \\
P_m(\vec x)=0,
\end{equation}
(where $\ …
4
votes
0
answers
227
views
To what extent is the Nash embedding not unique?
Consider a smooth Nash embedding, $f$, of a Riemannian manifold $Σ$ into Euclidean space $\mathbb R^n$. To what extent is this embedding not unique?
It is clear that the set of all such embeddings con …
2
votes
1
answer
333
views
Volume of submanifold as integral of delta-function
Let $M$ be an $n-m$ dimensional sub-manifold of $\mathbf R^n$ defined by the following set of equations:
\begin{equation}
f_1(\vec x)=0, \\
\vdots \\
f_m(\vec x)=0,
\end{equation}
(where $\vec x$ are …
1
vote
2
answers
261
views
Ricci scalar of submanifold of $\mathbf R^n$
Let $M$ be an $n-m$ dimensional sub-manifold of $\mathbf R^n$ defined by the following set of equations:
\begin{equation}
f_1(\vec x)=0, \\
\vdots \\
f_m(\vec x)=0,
\end{equation}
where $\vec x$ are c …
0
votes
Ricci scalar of submanifold of $\mathbf R^n$
Thanks Willie but I find your answer slightly hard to follow. The following is what I uncovered.
Given a basis $\{e_a\}$ (with $a=1,...,n-m$) for the tangent space $TM$ the second fundamental form, $\ …
2
votes
2
answers
380
views
Under what conditions can an orientable Riemannian 3-manifold be defined implicitly?
Under what conditions can an orientable Riemannian 3-manifold $\Sigma$ be defined implicitly?
What I mean by implicitly is that there exists a smooth function $f:\mathbb{R}^n\to \mathbb{R}^m$, such th …