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From the homotopy exact sequence of the fibration $$ \mathrm{SU}(2)\times \mathrm{SU}(2) \longrightarrow \mathrm{SU}(4)\longrightarrow \frac{\mathrm{SU}(4)}{\mathrm{SU}(2)\times \mathrm{SU}(2)} = Q $ …

Have you done any integration theory? (I assume you have, otherwise you wouldn't necessarily know what the deRham cohomology does for you.) The fastest proof I know is: Take a closed $k$-form $\o …

The answer is no. Suppose that $\alpha_1,\ldots,\alpha_g,\beta_1,\ldots,\beta_g$ were closed $1$-forms on $M$ such that their cohomology classes were a basis of $H^1(M)$ and they satisfied $\alpha_i …

In one sense, there are only Kähler examples, but, in another sense, there are non-Kähler examples of such $J$. Here is what I mean: Suppose that one has the data $(M,g,J)$ as defined in the questio …

What you are looking for nowadays goes by the name of the Rumin complex and is defined on any contact manifold. Moreover, there is a vast generalization of this that sometimes goes by the name of 'th …

The answer is 'no'. Already, it is not true for $M = \mathrm{Sp}(2)/\bigl(\mathrm{SU}(2)\times\mathrm{U}(1)\bigr)$, which is known to be diffeomorphic to $\mathbb{CP}^3$. To see this, note that the …

There is no retraction of $\mathrm{SO}(7)$ onto $\mathrm{G}_2$. If such a retraction $\rho:\mathrm{SO}(7) \to \mathrm{G}_2$ existed, then the composition $$ \mathrm{G}_2 \hookrightarrow \mathrm{SO}(7 …

Is this concrete enough? Recall that $\mathrm{SU}(3)$ fibers over $S^5$, with fibers equal to $\mathrm{SU}(2)$ and that this fibration is nontrivial. Let $S^1\subset \mathrm{SU}(2)$ be (any) subgrou …

They are homotopic when $n=2$, but not when $n>2$. Here is the argument: Let $A=(a_{ij})$. Then, the definitions of the two framings can be made more explicit as follows: For the first frame field …

If the normal bundle of $\Sigma$ in $M$ is orientable, then there always exists such a foliation. The idea is that one can construct a smooth function $f$ on $M$ such that $\Sigma$ is the set of zero …

I'm sorry if this is obvious to everyone, but I thought that it was worth mentioning: I don't know the answer to the question asked, but the answer to the easier question, "Does the space $\mathrm{E} …

The answer is 'no', the affine symplectic group cannot appear as a Lie subgroup of any compact Lie group. The reason is that the affine symplectic group contains $\mathrm{SL}(2,\mathbb{R})$ as a Lie …

If you just want an explicit realization the of outer automorphisms of $\mathrm{Spin}(8)$, here is one, assuming that you know about the algebra of octonions $\mathbb{O}$, the unique $8$-dimensional ( …

Here's an example to show that the infimum is not always attained: Consider the standard Hopf map $\pi:S^3\to S^2$, which is not null-homotopic, of course, so it follows that the area of the graph in …

There is a trivial construction that shows that the answer is 'yes' for all complex manifolds, not just those that admit an anti-holomorphic involution. Let $(M,J)$ be a (finite-dimensional) complex $ …