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Let $e$ denote the Euler class of the tautological line bundle $\gamma^n$ over $\mathbf{HP}^n$. My question is how to determine the pairing $$ \langle e,\ [\mathbf{HP}^n]\rangle$$ with $[\mathbf{HP}^n]$ denoting the fundamental class of $\mathbf{HP}^n$.

The only approach I could think of was to mirror the approach for $\mathbf{CP}^n$. For with $\mathbf{CP}^n$ I would use the fact that $$T\mathbf{CP}^n \oplus \varepsilon^1 \cong (\overline{\gamma^n})^{n+1}.$$ Pairing the Euler class of the tangent bundle yields the Euler characteristic (which is easy to compute), so I could then use Chern classes to give the answer. This was my plan for $\mathbf{HP}^n$ as well, until I was told (to my horror) that $T\mathbf{HP}^n$ does not have a complex-structure for any $n$.

Edit. I was vague about what I meant by "the" fundamental class of $\mathbf{HP}^n$; let me elaborate on this. I now realise that my question is actually poorly posed. What I am actually after is to express the Euler class of the tangent bundle in terms of the Euler class of the tautological bundle. However, this is itself ill posed as this too depends on a choice of orientation. The only case where I would know how to do this is in the case $n=1$, where this is simply $S^4$. But this would likely depend on the explicit diffeomorphism with $S^4$.

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  • $\begingroup$ I'm now confused as to the difference between $e$ and $e(\gamma^n)$.... $\endgroup$ May 5, 2020 at 16:30
  • $\begingroup$ @JasonDeVito You are completely right, and this rather embarrassing on my part... Apologies again. My question is fundamentally ill posed, and I now realise that my original problem still allowed for a freedom to pick the fundamental class. I have now also accepted your answer since it fully answers everything else very well. $\endgroup$
    – SvanN
    May 5, 2020 at 17:24
  • $\begingroup$ No need to apologize, I found the problem quite fun! $\endgroup$ May 5, 2020 at 17:43

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Short answer: The Euler class of the tautological bundle over $\mathbb{H}P^n$ is a generator of $H^4(\mathbb{H}P^n;\mathbb{Z})\cong \mathbb{Z}$.

I'm going to view $\mathbb{H}^n$ as a quaternionic "vector space" where the scalars multiply on the right. This allows me to keep linear transformations on the left.

Set $f:\mathbb{C}^{2n}\rightarrow \mathbb{H}^n$ to be the map which takes $(z_1, z_2,..., z_{2n})$ to $(z_1 + jz_2, z_3 + jz_4 ,..., z_{2n-1} + jz_{2n})$. Because of the placement of the $j$ and the fact that scalars multiply on the right, if we view $\mathbb{H}^n$ as a $\mathbb{C}$-vector space (by restricting $\mathbb{H}$ to $\mathbb{C}$), then $f$ is $\mathbb{C}$-linear.

I'll write $[z_1: .... :z_{n+1}]_\mathbb{C}$ for the equivalence classes the define $\mathbb{C}P^n$. That is, two nonzero $x,y\in \mathbb{C}^n$ are equivalent if there is a non-zero complex number $\lambda$ for which $x = y\lambda$. (Of course, $\mathbb{C}$ is commutative, so it doesn't matter what side $\lambda$ is on).

Likewise, I'll write $[p_1 : ... : p_{n+1}]_{\mathbb{H}}$ for quaternionic case. That is, I'll call $x,y\in \mathbb{H}P^n$ equivalent if $x = y\lambda$ for some non-zero $\lambda \in \mathbb{H}$. (Here, the location of $\lambda$ is important.)

Now, $f$ induces a map $g:\mathbb{C}P^{2n+1}\rightarrow \mathbb{H}P^n$ by $g([z_1:z_2:...:z_{2n+1}:z_{2n+1}]_{\mathbb{C}}) = [z_1 + jz_2: ... : z_{2n+1} + j z_{2n+2}]_\mathbb{H}$ which is well defined because the $\mathbb{C}$-scalar multiplication on $\mathbb{H}P^n$ is the restriction of the $\mathbb{H}$-scalar multiplication.

Now, let $\mathbb{H}\rightarrow E\xrightarrow{\pi} \mathbb{H}P^n$ be the $\mathbb{H}$-tautological bundle.

Proposition: Viewed as real bundles, the pull back bundle $g^\ast(E)$ splits as a sum of two copies of the $\mathbb{C}$-tautological bundle.

Proof: Let $T$ denote the tautological $\mathbb{C}$-bundle over $\mathbb{C}P^{2n+1}$. Define a map $\phi$ from $T\oplus T$ to $E$ by $\phi(z,v_1, v_2) = (g(z), f(v_1) + f(v_2)j)$.

Note that $\phi$ really has an image in $E$: because $v_1 \in z$ and $f$ is $\mathbb{C}$-linear, $f(v_1)\in g(z)$. Also, because $g(z)$ is a quaternionic line, $f(v_2)j \in g(z)$ iff $f(v_2) \in g(z)$. And, since $f$ is $\mathbb{C}$-linear, $f(v_2)\in g(z)$.

This shows that $\phi$ really does map into $E$. It's obviously $\mathbb{R}$-linear (because $\mathbb{R}\subseteq \mathbb{H}$ is in the center of $\mathbb{R}$), thought I don't think it's $\mathbb{C}$-linear.

Lastly, note that $\phi$ is injective: if $f(v_1) + f(v_2)j = 0$, then first note that $f(v_1)\in \mathbb{C}$ while $f(v_2)j\in \mathbb{C}j \bot \mathbb{C}$, so this implies that $f(v_1) = f(v_2) = 0$. Now, because $f$ itself is injective, this implies $v_1 = v_2 = 0$.

Thus, $T\oplus T$ fits in the same bundle diagram as $g^\ast(E)$, so they must be isomorphic. $\square$

Since the Euler class is natural, $g^\ast(e(E)) = e(g^\ast(E)) = e(T\oplus T) = e(T)\cup e(T)$. Since the top Chern class equals the Euler class for a complex bundle, $e(T) = a$, where $H^\ast(\mathbb{C}P^{2n+1}) \cong \mathbb{Z}[a]/a^{2n+2}$. It follows that $e(T\oplus T) = e(T)\cup e(T) = a^2$, so is a generator of $H^4(\mathbb{C}P^{2n+1})$.

Proposition: The map $g$ induces an isomorphism on $H^4$. In particular, the Euler class of $E$ is a generator of $H^4(\mathbb{H}P^n;\mathbb{Z})\cong \mathbb{Z})$.

Proof: The map $g$ is actually a fiber bundle map with fiber $S^2$. For example, the chain of subgroups $Sp(n)\times S^1\subseteq Sp(n)\times Sp(1)\subseteq Sp(n+1)$ gives a homogeneous fiber bundle $$ Sp(n)\times Sp(1)/ (Sp(n)\times S^1)\rightarrow Sp(n+1)/(Sp(n)\times S^1)\rightarrow Sp(n+1)/(Sp(n)\times Sp(1))$$ which is, up to diffeomoprhism, given by $S^2\rightarrow \mathbb{C}P^{2n+1} \xrightarrow{g} \mathbb{H}P^n$.

Now, the Euler class of this $S^2$-bundle must be trivial (since $H^3(\mathbb{H}P^n) = 0$), so the Gysin sequence splits into short exact sequences of the form $$ 0\rightarrow H^k(\mathbb{H}P^n)\xrightarrow:{g^\ast} H^k(\mathbb{C}P^{2n+1}) \rightarrow H^{k-2}(\mathbb{H}P^n)\rightarrow 0.$$ Now, if $k$ is a multiple of $4$ (so $H^k(\mathbb{H}P^n)$ is non-zero), then $k-2$ is not a mulitple of $4$, so $H^{k-2}(\mathbb{H}P^n) = 0$. Thus, this short exact sequence shows that $g^\ast$ is an isomorphism on degrees which are a multiple of $4$. $\square$.

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  • $\begingroup$ Thank you for your detailed answer! However, I'm not sure how it fully answers my question. For saying that $e$ is a generator of $H^4(\mathbf{HP}^n;\mathbf{Z})$ would only give that the pairing is $\pm 1$, i.e., it gives the pairing up to sign, right? This has been the trouble I'm having. $\endgroup$
    – SvanN
    Apr 29, 2020 at 6:36
  • $\begingroup$ Yes, it only gives the answer up to sign. When you write "the" fundamental class, how are you picking between the two? Once I know that, I may be able to resolve the sign ambiguity. $\endgroup$ Apr 29, 2020 at 11:48
  • $\begingroup$ I was indeed vague about this, my apologies. I have added a specification to the question. $\endgroup$
    – SvanN
    May 5, 2020 at 7:30

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