Let be a Riemannian manifold with metric , Levi-Civita connection , and curvature tensor . Define Apparently people sometimes refer to this as the curvature tensor, though it is probably not too confusing.

**Some algebra **

Recall also the following three identities, proved here:

- (Skew-symmetry)
- (Skew-symmetry)
- (Bianchi identity)

I claim now that there is a type of symmetry:

This is in fact a general algebraic lemma.

Lemma 1Let be a real vector space and let be a quadrilinear map satisfying the three bulleted identities. Then it satisfies the boxed one.

The proof is some slightly messy algebra, which I’ll only sketch. There is a geometrical way of thinking about this that Milnor presents in *Morse Theory*.

Let the Bianchi identity as written above be denoted by . If we add and use skew-symmetry several times, we obtain

Now using gives

and suitably interchanging all the variables gives the result.

**Sectional curvature **

Notation as above, for a 2-dimensional subspace , define the **sectional curvature** as

if form an orthonormal basis for .I now claim this is well-defined.

Proposition 2If span , then In particular, depends only on and is well-defined.

Indeed, write . Then this is a computation depending on the previous identities already proved. (Messy algebra and blogging do not mix well.) ** **

**Sectional curvature determines **

The sectional curvature actually encodes all the information contained in . Indeed, if we had two Riemannian metrics on the same manifold with curvature tensors with the same sectional curvature in all two-dimensional planes, then

for all .

Consider the difference . Then it satisfies all the four identities in the first section of this post, along with (by skew-symmetry again). Also by the fourth identity, is symmetric in , so the skew-symmetry just proved implies

In particular, we get another skew-symmetric identity:

Applying this again gives

and the Bianchi identity clearly gives that .

I should say something about the geometric interpretation about all this. If , then for a small neighborhood of is a surface in . Then the Gauss curvature of that surface is .

December 5, 2009 at 1:31 am

Dear,

Can you show me how to put a box outside your statement, for e.g. your lemmas, propositions, etc.

December 5, 2009 at 9:29 am

I use Luca Trevisan’s LaTeX to WordPress software; I type the posts first in LaTeX and the software apparently turns the theorem/lemma/proposition environments into boxes. The box style appears to vary based upon the theme of the blog.

I don’t know how to make a box directly.

December 5, 2009 at 11:23 am

Can you show me a simple example here?

December 5, 2009 at 12:18 pm

Here is the LaTeX code from a small piece of a previous post:

Back to ANT. Today, we tackle the case $e=1$. We work, for convenience, in the local case where all our DVRs are complete, and all our residue fields are perfect (e.g. finite).

I’ll just state these assumptions at the outset. Then, \textbf{unramified extensions} can be described fairly explicitly.

So fix DVRs $R, S$ with quotient fields $K,L$ and residue fields $\overline{K}, \overline{L}$. Recall that since $ef=n$, unramifiedness is equivalent to $f=n$, i.e.

\[ [\overline{L}:\overline{K}] = [L:K].\]

Now by the primitive element theorem (recall we assumed perfection of $\overline{K}$), we can write $\overline{L} = \overline{K}(\overline{\alpha})$ for some $\overline{\alpha} \in \overline{L}$.

The goal is to lift $\overline{\alpha}$ to a generator of $S$ over $R$.

Well, there is a polynomial $\overline{P}(X) \in \overline{K}[X]$ with $\overline{P}(\overline{a}) = 0$; we can choose $\overline{P}$ irreducible and thus of degree $n$. Lift $\overline{P}$ to $P(X) \in R[X]$ and $\overline{a}$ to $a’ \in R$; then of course $P(a’) \neq 0$ in general, but $P(a) \equiv 0 \mod \mathfrak{m}’$ if $\mathfrak{m}’$ is the maximal ideal in $S$, say lying over $\mathfrak{m} \subset R$. So, we use Hensel’s lemma to find $a$ reducing to $\overline{a}$ with $P(a)=0$—indeed $P'(a’)$ is a unit by separability of $[\overline{L}:\overline{K}]$.

I claim that $S = R[a]$. Indeed, let $T=R[a]$; this is an $R$-submodule of $S$, and

\[ \mathfrak{m} S + T = S \]

because of the fact that $S/\mathfrak{m’}$ is generated by $\alpha$ as a field over $k$.

Now Nakayama’s lemma WIKIPEDIA implies that $S=T$.

\begin{proposition}

Notation as above, if $L/K$ is unramified, then we can write $L=K(\alpha)$ for some $\alpha \in S$ with $S=R[\alpha]$; the irreducible monic polynomial $P$ satisfied by $\alpha$

remains irreducible upon reduction to $k$.

\end{proposition}

There is a converse as well:

\begin{proposition} If $L=K(\alpha)$ for $\alpha \in S$ whose monic irreducible $P$ remains irreducible upon reduction to $k$, then $L/K$ is unramified, and $S=R[\alpha]$.

\end{proposition}

Running latex2wp.py on this should give an output in html which when copied into the WordPress “New post” page and then posted will yield boxes.

December 5, 2009 at 2:45 pm

Thanks, got it 🙂 just use

and it really depends on the theme we are using 😦