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6

i think it is unlikely that this is true. Take a dumbbell type shape with rotational symmetry. Make the bar bit pinch in. The smallest circle is a totally geodesic submanifold and the invariant under rotations. Now deform the rest of the barbell whilst leaving a small neighbourhood of the smallest circle unchanged. If you make it lumpy enough there won't be ...


3

This simply cannot be true. A generic Riemannian manifold does not have any isometries, but still the image of any geodesic is a totally geodesic submanifold.


3

The problem is in the last line of your computation. You seem to be using $$ \langle \nabla_{e_i} (e^i \wedge \tau), \omega \rangle = - \langle e^i \wedge \tau, \nabla_{e_i} \omega \rangle . $$ All of your earlier formulas involving $e^i$ or $e_i$ are local formulas, which make sense only in the domain of the orthonormal frame $(e_i)$. But this one is not ...


2

There is no need to proceed via the Stiefel-Manifolds. You can directly realize the Grassmannians as homogeneous spaces. In the real case, this takes the form $G(k,n)=SO(n)/S(O(k)\times O(n-k))$. This corresponds to a so-called symmetric decomposition of the Lie algebra $\mathfrak{so}(n)$, thus making $G(k,n)$ into a compact symmetric space. (The Riemannian ...


2

This is a partial answer, mainly intended to point out a class of counterexamples. 1. If "isometry" means that $f : S \to S$ preserves the metric in the sense that $d_M(f(x), f(y)) = d_M(x,y)$ for points $x,y \in S$, then it seems that all you need is the assumption that $M$ is complete. In this case, of course, there's no need to assume that $S$ is a ...


2

The answer is no. The plane with taxicab distance $\|x\|=|x_1|+|x_2|$ is a counterexample. A $CAT(0)$ Finsler space is Riemannian. Indeed, CAT(0) condition is scale-invariant, therefore passes to tangent spaces. And the only CAT(0) normed spaces are those with an inner product norm. One reference is: Ptolemaic spaces and CAT(0) by Buckley, Falk, and ...


2

Regarding (1) a three manifold is conformally flat if and only if the Cotton tensor vanishes. The Cotton tensor is defined as a linear combination of covariant derivatives of the curvature tensor, and hence, in particular, always vanishes for constant curvature metrics. So yes, $h$ is by definition conformally flat. Regarding (2) a quick scan of the paper ...


2

Let $w=fdg$ where $f, g$ are $0$-forms Evaluate the first term $$dw(V, W)=df\wedge dg(V, W)=df(V)dg(W)-df(W)dg(V)=V(f)W(g)-V(g)W(f)$$ the second term $$Vω(W)−Wω(V)−ω([V,W])$$ $$=V(fdg(W))-W(fdg(V))-fdg([V,W])$$ $$=V(fW(g))-W(fV(g))-f(VW(g)-WV(g))$$ $$=V(f)W(g)-V(g)W(f)$$ are the same.


2

The Plücker embedding is an isometry of $G(k,\Bbb C^n)$ to its image in $\Bbb P(\Lambda^k\Bbb C^n)$ with the standard Fubini-Study metric. In the moving frames notation, for example, the Kähler form on $\Bbb P^N$ is given by $$\frac i2\sum_{j=1}^N \omega_{0\bar j}\wedge\overline\omega_{0\bar j},$$ where $\{f_0;f_1,\dots,f_N\}$ is a unitary frame at the ...


2

For $v\in T_pM$, we want to view $v$ as a directional derivative operator. That is, if $c$ is a curve with $c'(0)=v$, then $$ v(f) : =\frac{d}{dt} f\circ c(t) $$ Here we must show that $v(f)$ is independent of curve $c$. (1) Let $w:=(d\phi)^{-1}v $ Then if $c=\phi\circ \alpha,\ \alpha'(0)=w,$ $$ v(f)= \frac{d}{dt} f\circ c(t)=\frac{d}{dt} f\circ \phi\circ ...


2

This depends on what definition of the Ricci curvature you are working with. If you view it as a tensor, writing in index notation, the condition for a Kahler manifold to be Einstein is that $$R_{i \bar{j} } = \alpha g_{i\bar{j}} ,$$ for some constant $\alpha$. But one can also define the curvature form (also known as the Ricci form) to be $$ R = ...


2

You should just think about a simple closed curve $C$ in the plane, bounding a region $\Omega$. I don't know what your background is, but you should be able to show that if $\mathbf v$ has no zero inside the curve, then it must be somewhere tangent to the curve. In particular, if $\mathbf v$ is nowhere tangent, it always points out or it always points in, ...


2

In Riemannian manifold $(M,g)$, $X(f)$ implies that if $c(t)$ is a curve with $c'(0)=X$ then $$ X(f)=\frac{d}{dt}\bigg|_{t=0} f(c(t)) $$ So $$\frac{d}{dt} |\xi|^2= X |\xi|^2=X g(\xi,\xi) $$ In $\mathbb{R}^3$, $M$ is a surface. Then covariant derivative is defined as : $$ \nabla_X \xi = Proj_{T_{c(0)} M} \frac{d}{dt}\xi $$ where $Proj$ is a projection. So ...


1

The Gauss lemma states that for $X\in T_xM$ we have$$\langle d(\exp_x)_p(X),d(\exp_x)_p(p)\rangle=\langle X,p\rangle.$$It follows that the kernel of $d(\exp_x)_p$ is contained in the orthogonal complement of $p$, as you conjectured.


1

I don't think it's possible to have a manifold with the topology of $S^3$ and a Lorentzian metric. The corresponding maximally-symmetric Lorentzian spaces of constant (non-zero) curvature would be deSitter space (dS) and Anti-deSitter space (AdS). These would be the closest analogs to spheres.


1

I thought this question was interesting; here's what some internet research has turned up: 1. When $M$ is simply connected, the convexity radius of $M$ is $\infty$ iff $M$ has no focal points. On the one hand, if $M$ has no focal points then the argument linked to by Woodface in the comments above shows that all geodesic balls in $M$ are convex. On the ...


1

The Lie algebra of a compact, connected Lie group $G$ is abelian if and only if $G$ is a torus (that is, if and only if $G\cong \mathbb{S}^1\times\cdots\times \mathbb{S}^1$ is a finite product of circle groups); to prove this, note that the exponential map is the universal cover. So, abelian Lie subalgebras of the Lie algebra of a Lie group give rise to ...


1

I suppose you need to assume that the Gauss curvature of $S$ is positive. In this case, $N$ is a diffeo from $S$ onto the sphere. Otherwise, you should define what area of $N(A)$ means. In any case, if you use the Gauss-Bonnet theorem on $A$, then you obtain that $$ \int_A K=2\pi, $$ because the geodesic curvature of $\alpha$ vanishes, and the Euler ...


1

For vector fields $U, V, W$ on a ($C^1$) Riemannian manifold, we have the Leibniz Rule $$ W \langle U, V \rangle = \nabla_W \langle U, V \rangle = \langle \nabla_W U, V \rangle + \langle U, \nabla_W V \rangle \textrm{.} $$ Now, take $U := Z, V := Z, W := [X, Y]$. (In fact, this argument works for any connection $\nabla$ compatible with the metric.)


1

If $(M,g)$ is Riemannian manifold, the Levi-Cevita connection is the unique affine connection on $TM$ which is torsion-free and compatible w.r.t. $g$ : $\nabla_XY - \nabla_YX = [X,Y]$ $X \cdot\langle Y,Z\rangle = \langle \nabla_XY,Z\rangle + \langle Y,\nabla_XZ\rangle$ So, in particular, $$[X,Y]\cdot\langle Z,Z\rangle = \langle \nabla_{[X,Y]}Z,Z\rangle ...


1

$df(X):=X(f)$. And $$ \alpha:=gdf_1\wedge \cdots \wedge df_k \Rightarrow d(\alpha):= dg\wedge df_1\wedge \cdots \wedge df_k $$ Hence if $E_i$ is a local coordinate vector field and $E_i^\ast $ is a dual, then $$ df(E_i)=E_i(f) \Rightarrow df=E_i(f)E_i^\ast $$ Hence $$ d(df)=0 \Rightarrow d(E_i(f))\wedge E_i^\ast + E_i(f) dE_i^\ast =0 $$ Here $$ ...


1

Let $\partial_i$ denote the partial derivative wrt $y^i$ and set $X=\sum_i X_i\partial_i$. Then the $(n-1)$-form that we must differentiate is $i_XdV=\sum_j a_jdy^1\wedge\cdots\wedge\widehat{dy^j}\wedge\cdots\wedge dy^n$ with $$ ...


1

Note that $$ \partial_{\phi} = \cos\ \theta (-\sin\ \phi,\cos\ \phi,0),\ \partial_\theta = (-\sin\ \theta\cos\ \phi,-\sin\ \theta\sin\ \phi,\cos\ \theta)$$ (1) Since $f$ is a parametrization and since $|\partial_\theta| =1$, then $f(\theta, \phi_0)$ is a geodesic : $$ \nabla_{\partial \theta}\partial \theta =0 $$ (2) And note that $$ \partial_\phi \perp ...


1

This might be more a comment than an answer and I only adress your second question but here goes. The heat kernel will almost never depend only on the distance. For example, think of a manifold $M$ on which there are two distinct parts: a flat part (where you have Euclidean volume growth), and a very thin, cylindrical part (where the volume growth is much ...


1

Since you've done the rest of the work in your answer, I'll show that $DRc_g = $ id on im $\delta_g^*$. It ends up being just a consequence of the invariance of $Rc$ under the action of the diffeomorphism group on the metrics. For $X$ a vector field and $\varphi_t$ the 1-parameter group of diffeomorphisms generated by $X$, this invariance gives $$ ...


1

Notice that there is no Riemann tensor for an inner product space. Moreover, at one chosen point $p \in M$ the metric inner product in $T_p M$ can always be made Euclidean. In other words, the condition for a metric to be flat is local, not pointwise. To see what is going on, one may look at the Taylor expansion of the metric. In special ("geodesic" or ...


1

It's very common to write symmetric products using juxtaposition with no product symbol. But you should never omit the wedge symbol in a wedge product. Maybe you're getting confused with the notation for integrals, where, for example $\int f\,dx^1\wedge dx^2$ is defined to mean $\int f(x^1,x^2)\, dx^1dx^2$. The $dx^1dx^2$ in the latter expression is not a ...



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