Is true that $\left[\frac{\hat{A}(\mathbb HP^m)} { \hat{M}(\mathbb HP^m) }\right]_{4m} = 0$? There are two questions:


*

*How to prove that in general 
$[\hat{A}(\mathbb HP^m)]_{4m} = 0$
It is possible to verify it for low values of $m$.

*How to prove that in general
$\left[\frac{\hat{A}(\mathbb HP^m)}   {   \hat{M}(\mathbb HP^m) }\right]_{4m} = 0$
where $\hat{M}(\mathbb HP^m)$ is the Mayer class defined by
$\hat{M}(V) = \prod _{i=1}^{s}\cosh \left( \frac{y_{{i}}}{2} \right)$
with
$p(V) =\prod _{i=1}^{s}(1+{y_{i}}^2)$.
It is possible to verify it for low values of $m$.
 A: Since $HP^n$ is a spin manifold and since the homogeneous metric has positive scalar curvature, the $\hat{A}$-genus of $HP^n$ is zero by the index theorem and the Weitzenboeck-Lichenrowicz formula.
I cannot at the moment answer the question for the quotient by the Mayer class. But there is a direct computation for the $\hat{A}$-genus, which might help for the quotient as well. I learnt this method from Hirzebruch, when he was still around at the Max-Planck-Insitute.
In my lecture notes http://wwwmath.uni-muenster.de/u/jeber_02/skripten/mainfile.pdf, page 161, I discuss the computation of the basic characteristic classes of $HP^n$.
Consider the circle bundle $q: CP^{2n+1} \to HP^n$. $q$ induces an injection on cohomology. Therefore, it is enough to calculate that $(q^{\ast}\hat{A}(THP^n),[CP^{2n}])=0$.
One can show that $q^{\ast} THP^n \oplus H^{\otimes 2} = TCP^{2n+1}$, where $H$ is the Hopf bundle. Denote by $x \in H^2 (CP^{2n+1})$ the generator (first Chern class of $H$).
For a general even power series $F$ with associated multiplicative sequence, we therefore have $q^{\ast} F(THP^n) = F(x)^{2n+2}F(2x)^{-1}$. We have to determine the $2n$th coefficient and prove that it is zero for $n >0$, when $F$ is the power series of the $\hat{A}$-genus. The $2n$th coefficient is $\frac{1}{2 \pi i}\int F(x)^{2n+2}F(2x)^{-1} x^{-2n-1} dx$, integration is over a circle around $0$ in the complex plane.
Form the generating series 
$$
G(t):=\sum_{n \geq 0} \langle F(x)^{2n+2}F(2x)^{-1}; [CP^{2n}] \rangle t^{2n}
$$
Using the above formula for the coefficients and the sum formula for the geometric series, one sees that the generating series is the same as 
$$
\frac{1}{2 \pi i}\int \frac{F(x)^2}{F(2x)x}  \frac{1}{1- \frac{F(x)t}{x}} dx.
$$
For the $\hat{A}$-genus, $F(x) = \frac{x/2}{\sinh(x/2)}$. Perform the substitution $\sinh(x/2)=u$ in the above integral. Finally, you arrive at the integral
$$
\frac{1}{2 \pi i}\int \frac{1}{u-t/2} du.
$$
The value is independent of $t$, as long as $|t|$ is small enough. Therefore $G(t) \equiv 1$, and this proves the result.
I have not performed the first sanity check for this calculation (plug in the Hirzebruch $L$-class has to give the correct value for the signature of $HP^n$, namely $1$). 
A: I think that using the method of Professor Ebert is possible to prove the question 2 according with the following procedure.
For the expression ${\frac {\hat{A}}{\hat{M}}}$ the corresponding $F(x)$ is
$F(x) =  \frac{x/2}{\sinh(x/2) \cosh(x/2)} = \frac{x}{\sinh(x)} $
Then we have
$\frac{1}{2 \pi i}\int \frac{F(x)^2}{F(2x)x}  \frac{1}{1- \frac{F(x)t}{x}} dx = \frac{1}{2 \pi i}\int{\frac {\cosh \left( x \right) }{\sinh \left( x \right) -t}}dx$
With the substitution $u=\sinh \left( x \right)$, the integral is reduced to
$\frac{1}{2 \pi i}\int \frac{1}{u-t} du$
This last integral is independent of $t$, then $G(t) \equiv 1$ and then we have that
$[\frac{\hat{A}(\mathbb HP^m)}   {   \hat{M}(\mathbb HP^m) }]_{4m} = 0$
Do you agree?
A: It is possible to obtain a more direct proof using other idea due to Professor Ebert.
It is well known that 
$\hat{A} = \prod _ i{\frac {x_{{i}}/2}{\sinh \left( x_{{i}}/2 \right) }}= 1+\hat{A}_{1}+\hat{A}_{2}+\hat{A}_{3}+....$
where 
$\hat{A}_{1}=-\frac{1}{24}p_{{1}}$
$\hat{A}_{2} = -{\frac {1}{1440}}\,p_{{2}}+{\frac {7}{5760}}\,{p_{{1}}}^{2}$
$\hat{A}_{3} = {\frac {11}{241920}}\,p_{{1}}p_{{2}}-{\frac {1}{60480}}\,p_{{3}}-{
\frac {31}{967680}}\,{p_{{1}}}^{3} $
From other side
$ {\frac {\hat{A}}{\hat{M}}}= \prod _ i{\frac {x_{{i}}}{\sinh \left( x_{{i}} \right) }} = 1+ [{\frac {\hat{A}}{\hat{M}}}]_{1}+ [{\frac {\hat{A}}{\hat{M}}}]_{2}+[{\frac {\hat{A}}{\hat{M}}}]_{3}+....$
where
$[{\frac {\hat{A}}{\hat{M}}}]_{1} = -\frac{1}{6}p_{{1}}$
$[{\frac {\hat{A}}{\hat{M}}}]_{2} = {\frac {7}{360}}\,{p_{{1}}}^{2}-{\frac {1}{90}}\,p_{{2}} $ 
$[{\frac {\hat{A}}{\hat{M}}}]_{3} = {\frac {11}{3780}}\,p_{{1}}p_{{2}}-{\frac {31}{15120}}\,{p_{{1}}}^{3}-
{\frac {1}{945}}\,p_{{3}}
$
Then we have
$[{\frac {\hat{A}}{\hat{M}}}]_{1} = 4\hat{A}_{1}$
$[{\frac {\hat{A}}{\hat{M}}}]_{2} = 16\hat{A}_{2}$
$[{\frac {\hat{A}}{\hat{M}}}]_{3} = 64\hat{A}_{3}$
$[{\frac {\hat{A}}{\hat{M}}}]_{n}  = 4^{n}\hat{A}_{n}$
From the last equation we obtain that given that 
$[\hat{A}(\mathbb HP^m)]_{4m} = 0$
then
$[\frac{\hat{A}(\mathbb HP^m)}   {   \hat{M}(\mathbb HP^m) }]_{4m} = 0$.
Besides of this, the paragraph 

extracted from Baer, page 30;  can be simplified to the integrality of
$2^{l+\frac{n}{2}}\int_{X}\hat{A}_{\frac{n}{4}}(TX)$.
As an application of the last result it is possible to prove that $CP^4$ cannot be immersed in $R^{11}$.  Computing the last integral with $l=1$ and $n=8$ we obtain
$2^{5}\int_{CP^4}\hat{A}_{2}(TCP^4)= \frac{3}{4} $
which is not an integer.
In general for $CP^4$ the Mayer integral takes the form
$2^{l+4}\int_{CP^4}\hat{A}_{2}(TCP^4) = 3({2})^{l-3}$
Then this last integral indicates that  $CP^4$ cannot be immersed in $R^{10}, R^{11}, R^{12}, R^{13}$.  For $l\geq 3$ there is no topological obstruction to the existence of immersion of $CP^4$ in $R^{8+2l+1}$.
More in general, in the case of $CP^{2m}$ the Mayer integral is evaluated as
$2^{l+2m}\int_{CP^{2m}}\hat{A}_{m}(TCP^{2m}) = (-1)^m2^{l-2m}{2\,m\choose m}$.
