User michael albanese - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T23:20:12Z http://mathoverflow.net/feeds/user/21564 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/125582/alternative-almost-complex-structures Alternative Almost Complex Structures Michael Albanese 2013-03-26T02:49:55Z 2013-03-26T08:21:00Z <p>Originally posted on <a href="http://math.stackexchange.com/q/332667/39599" rel="nofollow">Maths Stack Exchange</a>.</p> <hr> <p>Let $V$ be a real vector space. An <em>almost complex structure</em> on $V$ is a map $J : V \to V$ such that $J^2 = -\mathrm{id}_V$. An almost complex structure gives $V$ the structure of a complex vector space by defining $(a+bi)v = av + bJ(v)$. The idea behind this definition is that $J$ represents multiplication by $i$. What if you instead consider multiplication by other complex numbers?</p> <p>Let $n > 2$ be a fixed positive integer and let $J : V \to V$ be such that $J^n = -\mathrm{id}_V$ which I will refer to as an <em>alternative almost complex structure</em>. One can regard $J$ as trying to capture multiplication by $\zeta = e^{\frac{\pi i}{n}}$. In particular, as ${1, \zeta}$ is linearly independent over $\mathbb{R}$, it is a basis for $\mathbb{C}$ as a real vector space. One can then endow $V$ with a complex structure by defining $(a + b\zeta)v = av + bJ(v)$.</p> <blockquote> <p>Do alternative almost complex structures give rise to the same results as the standard almost complex structures? In particular, in the case where we extend them to bundle endomorphisms of the tangent bundle of a manifold. </p> <p>If not, what fails? If so, is dealing with almost complex structures easier than dealing with alternative almost complex structures?</p> </blockquote> <hr> <p>As Henry T. Horton asks about in a comment to the original post, I am interested in the integrability of alternative almost complex structures but also whether there are any difficulties when combining with other structures such as a symplectic form or a Riemannian metric. </p> http://mathoverflow.net/questions/98836/question-about-an-estimate-in-hormanders-proof-of-cartans-theorem-b Question about an estimate in Hörmander's proof of Cartan's Theorem B Michael Albanese 2012-06-05T02:38:49Z 2013-01-13T11:47:55Z <p>I have been working through the proof of Cartan's Theorem B that Hörmander gives in his book 'Introduction to Complex Analysis in Several Variables'. When I began, I skipped over some of the initial steps so that I could get to a specific calculation that my supervisor wanted me to look at. Now, looking back at the setup, I realise that I don't understand all of the details.</p> <p>In Chapter 4, Hörmander introduces linear, closed, densely defined operators $T : L^2_{(p,q)}(\Omega, \varphi_1) \to L^2_{(p,q+1)}(\Omega, \varphi_2)$ and $S : L^2_{(p,q+1)}(\Omega, \varphi_2) \to L^2_{(p,q+2)}(\Omega, \varphi_3)$ which are defined by $\overline{\partial}$.</p> <p>Some functional analysis shows that it is enough to prove that there is a positive constant $C$ such that $\|f\|^2 \leq C(\|T^*f\|^2 + \|Sf\|^2)$ for all $f \in D_{T^*}\cap D_S$. An argument is then given to show that $D_{(p,q+1)}(\Omega)$ is dense in $D_{T^*}\cap D_S$ with respect to the graph norm $f \mapsto \|f\| + \|T^*f\| + \|Sf\|$, where $D_{(p,q+1)}(\Omega)$ denotes the smooth compactly supported $(p,q+1)$ forms. The proof of this fact is where my troubles begin.</p> <p>Hörmander shows that for suitable weights, and a sequence of compactly supported functions <code>$(\eta_{\nu})_{\nu \in \mathbb{N}}$</code> with $0 \leq \eta_{\nu} \leq 1$ and $\eta_{\nu} = 1$ on any compact subset of $\Omega$ when $\nu$ is large (which satisfy an appropriate bound on $|\bar{\partial}\eta_{\nu}|$), we have $\|\eta_{\nu}f - f\|_{\varphi_2} \to 0$, $\|S(\eta_{\nu}f) - \eta_{\nu}Sf\|_{\varphi_3} \to 0$, and $\|T^*(\eta_{\nu}f) - \eta_{\nu}T^*f\|_{\varphi_1} \to 0$. I can understand why the first two are true, but not the third.</p> <p>Hörmander shows that $\eta_{\nu}f \in D_{T^*}$. From there I can see how he gets, for $u \in D_T$, $|(T^*(\eta_{\nu}f) - \eta_{\nu}T^*f, u)_{\varphi_1}| \leq \int|f|e^{-\varphi_2/2}|u|e^{-\varphi_1/2}d\lambda$, but after this inequality, he states </p> <blockquote> <p>$\dots$ which implies the bound $|T^*(\eta_{\nu}f) - \eta_{\nu}T^*f|^2e^{-\varphi_1} \leq |f|^2e^{-\varphi_2}$.</p> </blockquote> <p>I don't see how this follows. How does Hörmander obtain this (pointwise) estimate? At the moment, the best I’ve got is a messy measure theoretic argument that I’m not even sure is correct. Any help would be much appreciated.</p> http://mathoverflow.net/questions/99912/almost-complex-structure-approach-to-deformation-of-compact-complex-manifolds Almost Complex Structure approach to Deformation of Compact Complex Manifolds Michael Albanese 2012-06-18T15:56:57Z 2013-01-10T08:44:32Z <p>I don't know much about the deformation of compact complex manifolds, I've only read chapter 6 of Huybrechts' book <i>Complex Geometry: An Introduction</i>. There are two parts to this chapter. The second goes through the standard approach, that is, considering a family of compact complex manifolds as a proper holomorphic submersion between two connected complex manifolds. My question is about the approach taken in the first section, which I will briefly outline.</p> <p>One can instead consider a deformation of complex <i>structures</i> on a fixed smooth manifold, as opposed to deformations of complex manifolds &ndash; by Ehresmann's result, a deformation over a connected base is nothing but a deformation of complex structure on a fixed smooth manifold. This point of view is difficult to work with because a complex structure is a complicated object, so we instead consider almost complex structures &ndash; by the Newlander-Niremberg Theorem, complex structures correspond to <i>integrable</i> almost complex structures.</p> <p>Fix a smooth even-dimensional manifold $M$. Now Huybrechts considers a continuous family of almost complex structures $I(t)$. He does not say where $t$ comes from, but I have interpreted it to be an open neighbourhood of $0$ in $\mathbb{C}$. Now, let $I(0) = I$. The complexified tangent bundle to $M$ splits with respect to $I$. That is, $TM\otimes_{\mathbb{R}}\mathbb{C} = T^{1,0}M\oplus T^{0,1}M$. But this is true of each almost complex structure $I(t)$. Denote the corresponding decompositions by $TM\otimes_{\mathbb{R}}\mathbb{C} = T^{1,0}M_t\oplus T^{0,1}M_t$ &ndash; this is deliberately suggestive notation; we can consider the compact (soon-to-be) complex manifold $(M, I(t))$ as the fibre of a complex family over a point $t$ in the base.</p> <p>For small $t$, we can encode the given information by a map $\phi(t) : T^{0,1}M \to T^{1,0}M$ where, for $v \in T^{0,1}M$, $v + \phi(t)v \in T^{0,1}M_t$. Huybrechts then says:</p> <blockquote> <p>Explicitly, one has $\phi(t) = -\text{pr}_{T^{1,0}M_t}\circ j$, where $j : T^{0,1}M \subset TM\otimes_{\mathbb{R}}\mathbb{C}$ and $\text{pr}_{T^{1,0}M_t} : TM\otimes_{\mathbb{R}}\mathbb{C} \to T^{1,0}M_t$ are the natural inclusion respectively projection.</p> </blockquote> <p>According to this, the codomain of $\phi(t)$ is $T^{1,0}M_t$, not $T^{1,0}M$. <b>Is this a typo or am I missing something?</b> <em>Added later:</em> As Peter Dalakov points out in his answer, it is a typo.</p> <p>Anyway, Huybrechts continues with this approach. Enforcing the integrability condition $[T^{0,1}M_t, T^{0,1}M_t] \subset T^{0,1}M_t$ ensures that each almost complex structure is induced by a complex structure. Under the assumption that $I$ is integrable, $[T^{0,1}M_t, T^{0,1}M_t] \subset T^{0,1}M_t$ is equivalent to the Maurer-Cartan equation $\bar{\partial}\phi(t) + [\phi(t), \phi(t)] = 0$, where $\bar{\partial}$ is the natural operator on the holomorphic vector bundle $T^{1,0}M$, and $[\bullet, \bullet]$ is an extension of the Lie bracket.</p> <p>I like this approach because if you take a power series $\sum_{t=0}^{\infty}\phi_it^i$ of $\phi(t)$ you can deduce:</p> <ol> <li>$\phi_1$ defines the Kodaira-Spencer class of the deformation;</li> <li>all the obstructions to finding the coefficients $\phi_i$ lie in $H^2(M, T^{1,0}M)$.</li> </ol> <p><b>Does anyone know of some other places where I would be able to learn about this approach, or is there some reason why this approach is not that common?</b></p> <p>Just for the record, I have looked at Kodaira's <i>Complex Manifolds and Deformation of Complex Structures</i>, but I haven't been able to find anything resembling the above.</p> http://mathoverflow.net/questions/108528/choosing-a-kahler-metric-which-restricts-the-norms-of-some-forms Choosing a Kähler metric which restricts the norms of some forms Michael Albanese 2012-10-01T08:28:54Z 2013-01-06T05:23:07Z <p>Let $X$ be a non-compact complex manifold of Kähler type (i.e. there exists a Kähler metric on $X$ but it hasn't been endowed with one). For each $i \in \mathbb{N}$, let $f_i$ be a smooth function $X \to [0, 1]$ such that for every compact set $K \subset X$, there exists $N \in \mathbb{N}$ such that $f_i|_K = 1$ for all $i \geq N$.</p> <blockquote> <p>Is it always possible to choose a Kähler metric on $X$ such that, for all $i$, $\|\bar{\partial}f_i\| \leq 1$ with respect to the norm on $\Omega^{0,1}(X)$ induced by the metric?</p> </blockquote> <p>If the answer is no, can we do any better than a generic hermitian metric? That is, can we find a locally conformally Kähler, Hermitian-Einstein, or some other special metric which will give $\|\bar{\partial}f_i\| \leq 1$?</p> http://mathoverflow.net/questions/107168/weitzenbock-identities Weitzenböck Identities Michael Albanese 2012-09-14T12:07:44Z 2013-01-06T03:30:27Z <p>I asked this question at <a href="http://math.stackexchange.com/q/194749/39599" rel="nofollow">Maths Stack Exchange</a>, but I haven't received any replies yet (I'm not sure how long I should wait before it is acceptable to ask here, assuming there is such a period of time).</p> <hr> <p>The Wikipedia page for <a href="http://en.wikipedia.org/wiki/Weitzenb%C3%B6ck_identity" rel="nofollow">Weitzenböck identities</a> is explicitly example based. I am looking for a reference which takes a more rigorous approach (as well as a discussion of the Bochner technique). In particular, I am interested in references which focus on these identities in complex geometry.</p> <p>I have already consulted Griffiths &amp; Harris which is mentioned in the article, but it only contains one example. Berger's <em>A Panoramic View of Riemannian Geometry</em> doesn't have much more.</p> <p>My interest in Weitzenböck identities has been motivated by a question arising from the following theorem:</p> <blockquote> <p>Let $X$ be a Kähler manifold and $E$ a hermitian holomorphic vector bundle with Chern connection $\nabla$. Then for the Laplacians $\Delta_{\bar{\partial}} = \bar{\partial}\bar{\partial}^* + \bar{\partial}^*\bar{\partial}$, $\Delta_{\partial} = \partial\partial^* + \partial^*\partial$, we have $\Delta_{\bar{\partial}} = \Delta_{\partial} + [iF_{\nabla}, \Lambda]$.</p> </blockquote> <p>Is $\Delta_{\bar{\partial}} = \Delta_{\partial} + [iF_{\nabla}, \Lambda]$ an example of a Weitzenböck identity?</p> http://mathoverflow.net/questions/116559/where-do-the-kahler-identities-first-appear Where do the Kähler Identities first appear? Michael Albanese 2012-12-17T02:56:15Z 2012-12-17T08:54:40Z <p>The <a href="http://mathworld.wolfram.com/KaehlerIdentities.html" rel="nofollow">Kähler identities</a> (sometimes known as the Hodge identities) are an important collection of relationships between operators on the exterior algebra of a Kähler manifold. These relationships generalise to hermitian manifolds, sections of hermitian holomorphic vector bundles, and many other situations. </p> <p>I know that the notion of a Kähler metric was introduced by Kähler himself in 1933 and that the Kähler identities were first generalised to hermitian manifolds by Demailly in 1985 (although he mentions that the ideas were present in a paper by Griffiths in 1966). </p> <blockquote> <p>Can anyone fill in the historical gap and tell me where or when the Kähler identities first appeared?</p> </blockquote> http://mathoverflow.net/questions/100530/does-equality-of-laplacians-imply-kahler/116146#116146 Answer by Michael Albanese for Does equality of Laplacians imply Kähler? Michael Albanese 2012-12-12T05:39:03Z 2012-12-12T05:39:03Z <p>In addition to the papers mentioned by YangMills, there is also the <a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=MR&amp;s1=0205205&amp;loc=fromreflist" rel="nofollow">earlier paper</a> by A. W. Adler which shows that if $\Delta = 2\Delta_{\bar{\partial}}$ on a hermitian manifold, then it is Kähler.</p> http://mathoverflow.net/questions/100530/does-equality-of-laplacians-imply-kahler Does equality of Laplacians imply Kähler? Michael Albanese 2012-06-24T16:27:15Z 2012-12-12T05:39:03Z <p>This question follows on from <a href="http://mathoverflow.net/questions/21315/non-kahler-manifolds-where-the-different-laplacians-are-compatible" rel="nofollow">this one</a>. </p> <p>Let $(X, \omega)$ be a Hermitian manifold and define the Laplacians $\Delta_{\partial} = \partial\partial^* + \partial^*\partial$ and $\Delta_{\bar{\partial}} = \bar{\partial}\bar{\partial}^* + \bar{\partial}^*\bar{\partial}$. </p> <p>If $(X, \omega)$ is a K&auml;hler manifold, that is $d\omega = 0$ (or equivalently $\partial\omega = 0$ or $\bar{\partial}\omega = 0$), we have $\Delta_{\bar{\partial}} = \Delta_{\partial}$. </p> <p>More generally, on any Hermitian manifold we have $\Delta_{\bar{\partial}} = \Delta_{\partial} + [\partial, [\Lambda_{\partial\omega}, L]] - [\bar{\partial}, [\Lambda_{\bar{\partial}\omega}, L]]$ where:</p> <ul> <li> $[\bullet, \bullet]$ is the graded commutator;</li> <li> $\Lambda_{\partial\omega}$ and $\Lambda_{\bar{\partial}\omega}$ are the adjoints of wedging with the forms $\partial\omega$ and $\bar{\partial}\omega$ respectively; and</li> <li> $L$ is the Lefschetz operator, that is, wedging with $\omega$.</li> </ul> <p>It is clear how the additional terms relating the Laplacians in the Hermitian case vanish if the metric is K&auml;hler ($\partial\omega = 0$ and $\bar{\partial}\omega = 0$, so $\Lambda_{\partial\omega}$ and $\Lambda_{\bar{\partial}\omega}$ are both zero). What about the converse? That is:</p> <blockquote> <p>If $\Delta_{\bar{\partial}} = \Delta_{\partial}$ on a Hermitian manifold $(X, \omega)$, is it necessarily K&auml;hler?</p> </blockquote> <hr> <p>The accepted answer in the linked question refers to <i>balanced</i> manifolds. These are manifolds with the property that $\Delta_{\bar{\partial}}f = \Delta_{\partial}f$ for any smooth <i>function</i> $f$. Not all such manifolds are K&auml;hler. The above question is stronger as it requires equality for all smooth <i>forms</i>.</p> http://mathoverflow.net/questions/106398/non-compact-kahler-manifolds-which-admit-a-positive-line-bundle Non-compact Kähler manifolds which admit a positive line bundle Michael Albanese 2012-09-05T04:33:23Z 2012-09-05T04:33:23Z <p>A complex manifold which admits a positive line bundle is automatically Kähler. Furthermore, if the manifold is compact, then it is projective by the Kodaira Embedding Theorem. In particular, not every compact Kähler manifold can admit a positive line bundle. What about in the non-compact case? That is:</p> <blockquote> <p>Are there any restrictions as to which non-compact Kähler manifolds can admit a positive line bundle?</p> </blockquote> http://mathoverflow.net/questions/102116/riemannian-metrics-as-sections-of-a-vector-bundle Riemannian metrics as sections of a vector bundle Michael Albanese 2012-07-13T08:29:27Z 2012-07-13T11:18:05Z <p>Let $\pi : E \to M$ be a smooth vector bundle. A Riemannian metric on $E$ can be regarded as a global section of the vector bundle $(E\otimes E)^{\ast}$, or more specifically, the subbundle $(S^2E)^{\ast} \subset (E\otimes E)^{\ast}$. However, not every global section $s$ corresponds to a Riemannian metric, as there is no gaurantee that $s(x) : S^2E_x \to \mathbb{R}$, when viewed as a symmetric map $E_x \times E_x \to \mathbb{R}$, is positive-definite. So my question is whether we can form a bundle such that positive-definiteness is automatic. More succinctly:</p> <blockquote> <p>Is there a vector bundle $F$ associated to $E$ such that the global sections of $F$ are precisely the Riemannian metrics on $E$?</p> </blockquote> <p>If so, can we characterise all Hermitian metrics on a complex vector bundle in a similar way?</p> http://mathoverflow.net/questions/100906/hermitian-christoffel-symbols Hermitian Christoffel Symbols Michael Albanese 2012-06-29T04:31:43Z 2012-07-04T23:37:58Z <blockquote> <p>Does anyone know of some good references for computing Christoffel symbols for Hermitian metrics?</p> </blockquote> <p>A quick Google search turns up <a href="http://arxiv.org/abs/1011.0207" rel="nofollow">this</a>. The following formula appears on page 4:</p> <p>$$\Gamma_{AB}^C = \frac{1}{2}h^{CE}\left(\frac{\partial h_{AE}}{\partial z^B} + \frac{\partial h_{BE}}{\partial z^A} - \frac{\partial h_{AB}}{\partial z^E}\right)$$</p> <p>where $A, B, C, E \in$ {$1, \dots, n, \bar{1}, \dots, \bar{n}$} and $z^{\bar{i}} = \bar{z}^i$. From this they get</p> <p>$$\Gamma_{\bar{i}j}^k = \frac{1}{2}h^{k\bar{l}}\left(\frac{\partial h_{j\bar{l}}}{\partial \bar{z}^i} - \frac{\partial h_{j\bar{i}}}{\partial \bar{z}^l}\right)$$</p> <p>How do they obtain this? Are they regarding $h$ as a map $(T^{1,0}M\oplus T^{0,1}M) \times (T^{1,0}M\oplus T^{0,1}M) \to \mathbb{C}$ where $h_{ab} = 0$, $h_{\bar{a}\bar{b}} = 0$, and $h_{\bar{a}b} = \overline{h_{b\bar{a}}}$? Even if they do, I don't see how they get the second term.</p> <p>Everything else I have found deals only with K&auml;hler metrics, in which case $\Gamma_{ab}^c$ and $\Gamma_{\bar{a}\bar{b}}^{\bar{c}}$ are the only non-trivial symbols.</p> <blockquote> <p>More generally, are there any treatments of Hermitian geometry which take this coordinate approach (as is common in Riemannian geometry texts)?</p> </blockquote> http://mathoverflow.net/questions/88996/non-compact-complex-surfaces-which-are-not-kahler Non-compact complex surfaces which are not Kähler Michael Albanese 2012-02-20T05:12:37Z 2012-06-27T06:52:46Z <p>Not every complex manifold is a Kähler manifold (i.e. a manifold which can be equipped with a Kähler metric). All Riemann surfaces are Kähler, but in dimension two and above, at least for compact manifolds, there is a necessary topological condition (i.e. the odd Betti numbers are even). This condition is also sufficient in dimension two, but not in higher dimensions. Therefore the task of finding examples of compact complex manifolds which are not Kähler is reduced to topological considerations.</p> <p>In the non-compact setting, we can also find such manifolds. For example, let $H$ be a Hopf surface, which is a compact complex surface which is not Kähler. Then for $k > 0$, $M_{k+2} = H\times\mathbb{C}^k$ is a non-compact complex manifold which is not Kähler - any submanifold of a Kähler manifold is Kähler, and $H$ is a submanifold of $M_{k+2}$. This generates examples in dimensions three and above. So I ask the following question:</p> <blockquote> <p>Does anyone know of some (easy) explicit examples of non-compact complex surfaces which are not Kähler?</p> </blockquote> http://mathoverflow.net/questions/99787/form-of-even-perfect-numbers/99789#99789 Answer by Michael Albanese for Form of even perfect numbers Michael Albanese 2012-06-16T13:43:14Z 2012-06-18T14:48:37Z <p>Every even perfect number is of the form $2^{p-1}(2^p - 1)$ where $2^p - 1$ is a (Mersenne) prime. Note that $p$ must be prime &ndash; if $p = ab$ with $a, b > 1$ then</p> <p>$$2^p - 1 = 2^{ab} - 1 = (2^a)^b - 1 = (2^a - 1)(1 + 2^a + 2^{2a} + \dots\ + 2^{a(b-1)}).$$</p> <p>If $p = 2$, we obtain the first perfect number $6$ which satisfies $6 \equiv 1\ (\text{mod 5})$. Every other prime is odd, so let $p = 2k + 1$. Then </p> <p>$$2^{p-1}(2^p - 1) = 2^{2k}(2^{2k+1} - 1) = 2.2^{4k} - 2^{2k} = 2.16^k - 4^k \equiv 2 - (-1)^k\ (\text{mod 5}).$$</p> <p>So, for $p = 2k + 1$,</p> <p>$$2^{p-1}(2^p - 1) \equiv \begin{cases} 1 \ (\text{mod 5}) &amp; \text{if }k\text{ is even}\newline 3 \ (\text{mod 5}) &amp; \text{if }k\text{ is odd}. \end{cases}$$</p> http://mathoverflow.net/questions/128903/expected-edit-distance Comment by Michael Albanese Michael Albanese 2013-04-27T17:15:09Z 2013-04-27T17:15:09Z Trying to understand the distance: $E(01010, 00100) = 2$ because $01010 \to 001010$ (insert $0$ inbetween first and second symbol) and $001010 \to 00100$ (remove the second $1$)? http://mathoverflow.net/questions/126158/a-mother-of-all-groups-what-kind-of-structures-have-mother-of-alls Comment by Michael Albanese Michael Albanese 2013-04-01T08:10:03Z 2013-04-01T08:10:03Z At least for finite groups there is this: <a href="http://en.wikipedia.org/wiki/Hall%27s_universal_group" rel="nofollow">en.wikipedia.org/wiki/Hall%27s_universal_group</a>. http://mathoverflow.net/questions/118506/transformation-of-symplectic-structure-by-a-matrix Comment by Michael Albanese Michael Albanese 2013-01-10T08:25:47Z 2013-01-10T08:25:47Z This isn't a question. http://mathoverflow.net/questions/118027/powers-of-a-degenerate-form Comment by Michael Albanese Michael Albanese 2013-01-04T08:49:22Z 2013-01-04T08:49:22Z Well, I don't know precisely what the etiquette is, but I think it is considered polite to let users of both sites know that the question has been posted elsewhere. That way, if someone here on MO knows the answer to the question, they can check MSE first to see if it has already been answered there, potentially saving them a lot of time and effort. http://mathoverflow.net/questions/118027/powers-of-a-degenerate-form Comment by Michael Albanese Michael Albanese 2013-01-04T08:29:39Z 2013-01-04T08:29:39Z Did you post this question on MSE (<a href="http://math.stackexchange.com/questions/270147/degenerate-2-form-and-its-powes" rel="nofollow" title="degenerate 2 form and its powes">math.stackexchange.com/questions/270147/&hellip;</a>) under a different name? If you post the same question on both sites, you should say so. http://mathoverflow.net/questions/29964/which-spheres-are-complex-manifolds/29977#29977 Comment by Michael Albanese Michael Albanese 2012-12-25T07:01:26Z 2012-12-25T07:01:26Z As an almost complex structure on $X$ endows $TX$ with the structure of a complex vector bundle, doesn't your argument show that these spheres don't even admit almost complex structures, let alone integrable ones? http://mathoverflow.net/questions/116559/where-do-the-kahler-identities-first-appear/116560#116560 Comment by Michael Albanese Michael Albanese 2012-12-19T03:29:25Z 2012-12-19T03:29:25Z I've had a look at Hodge's paper, but I can't see where exactly the K&#228;hler identities appear. http://mathoverflow.net/questions/116559/where-do-the-kahler-identities-first-appear/116560#116560 Comment by Michael Albanese Michael Albanese 2012-12-17T20:01:15Z 2012-12-17T20:01:15Z May I ask, how did you find this out? http://mathoverflow.net/questions/116559/where-do-the-kahler-identities-first-appear/116577#116577 Comment by Michael Albanese Michael Albanese 2012-12-17T09:39:21Z 2012-12-17T09:39:21Z In 'Eugenio Calabi and K&#228;hler Metrics' by Bourguinon, he attributes the introduction of the notion of a K&#228;hler metric to K&#228;hler's 1933 paper, but makes a parenthetical remark listing the paper you mention, as well as a paper by Schouten from 1929, as &quot;earlier attempts&quot;. I'm not sure why these two papers are only listed as attempts; personally, I have not had a look at either of them. http://mathoverflow.net/questions/108528/choosing-a-kahler-metric-which-restricts-the-norms-of-some-forms Comment by Michael Albanese Michael Albanese 2012-10-06T09:55:29Z 2012-10-06T09:55:29Z @BS: Thanks, I didn't think of this. I've restricted the $f_i$ to be converging to the constant function $1$. http://mathoverflow.net/questions/108528/choosing-a-kahler-metric-which-restricts-the-norms-of-some-forms Comment by Michael Albanese Michael Albanese 2012-10-06T08:06:54Z 2012-10-06T08:06:54Z @BS: Thanks for your comment. I have modified the question so that it more accurately reflects the situation I have come across. I think the hypothesis you proposed is now satisfied. http://mathoverflow.net/questions/108170/hyperreal-finitely-additive-measure-on-0-1-assigning-b-a-to-a-b-or-a-b Comment by Michael Albanese Michael Albanese 2012-09-26T22:36:05Z 2012-09-26T22:36:05Z @Alexander: In the title, it should be $(a, b]$ not $(b, a]$. http://mathoverflow.net/questions/107168/weitzenbock-identities/107189#107189 Comment by Michael Albanese Michael Albanese 2012-09-17T00:50:33Z 2012-09-17T00:50:33Z Sorry if I am missing something, but on page 530 of your notes you define the Hodge-Dolbeault operator to be $\bar{\partial} + \bar{\partial}^*$ but above you define it to be $\mathscr{D} = \frac{1}{\sqrt{2}}(\partial + \bar{\partial})$. Furthermore, isn't $\mathscr{D}(\Omega^{0,even}(M)) \subseteq \Omega^{1,even}(M)\oplus\Omega^{0,odd}(M)$? http://mathoverflow.net/questions/107168/weitzenbock-identities/107189#107189 Comment by Michael Albanese Michael Albanese 2012-09-16T14:29:27Z 2012-09-16T14:29:27Z Also, in Chapter 11 of your notes, you say that a Dirac operator is one which squares to be a generalised Laplacian which (I'm pretty sure) does not agree with what you have written. Using this definition instead and setting $D_{1, 0} = \partial + \partial^*$ and $D_{0,1} = \bar{\partial} + \bar{\partial}^*$ we have $D_{1,0}^2 = \Delta_{\partial}$ and $D_{0,1}^2 = \Delta_{\bar{\partial}}$. Is that what you meant? These are just the complex anologues of your Example 11.1.5 on the Hodge-de Rham operator $d + d^*$. http://mathoverflow.net/questions/107168/weitzenbock-identities/107189#107189 Comment by Michael Albanese Michael Albanese 2012-09-16T14:25:41Z 2012-09-16T14:25:41Z Thank you for your wonderful answer. I am not sure how the Hodge-Dolbeault operator (call it $D$ and ignore the constant factor of $\frac{1}{\sqrt{2}}$) fits in with the Laplacians $\Delta_{\partial}$ and $\Delta_{\bar{\partial}}$. You say that you can obtain a generalised Laplacian using $DD^*$ or $D^*D$, but neither of these give $\Delta_{\partial}$ or $\Delta_{\bar{\partial}}$. Instead, $DD^* + D^*D = \Delta_{\partial} + \Delta_{\bar{\partial}}$.