Since we homo sapiens (and all other species) are continuously evolving, are humans undergoing speciation based on our environment? OR has there been any difference across people that could say speciation has already happened?  PS: I am asking this question because I have recently seen a documentary in which some Chinese anthropologists claim that Chinese people are a different species altogether. Ref:  (Section )
After my online research on the subject, I learnt that, biologically speaking, many scientists believe that there is no such thing as a race. Homo sapiens as a species is only 200,000 years old, which has not allowed for any significant genetic diversification yet, and our DNA is 99.99% similar. I've read statements that there can be more genetic variation inside a racial group than between different racial groups, meaning that, for example, two individuals from the same "race" can have less in common with each other than with an individual from another "race".  :     Scientists consider biological essentialism obsolete, and generally discourage racial explanations for collective differentiation in both physical and behavioral traits     Q1: If Homo sapiens has no races (according to biologists), why are we so different morphologically? (hair/eyes/skin colour and even athletic performance seem to differ between human populations)  Q2: Is it common for other species too, when genetically close populations have very different morphological traits? Are there any other mammal or animal species that exhibit biological diversity comparable to human diversity, and how do taxonomists treat these species? (excluding intentionally bred domestic species to keep the comparison fair)    The question has been paraphrased to emphasize that it is the biological debate that is in question, not the sociopolitical. I.e., why is there no consensus in evidence and opinions of scientists?
How to get Maven to run war:exploded but not war:war I have a Maven pom that uses &lt;packaging&gt;war&lt;/packaging&gt;.  But actually, I don't want build the war-file, I just want all the dependent jars collected and a full deployment directory created.  So I'm running the war:exploded goal to generate the deploy directory:  &lt;plugin&gt;     &lt;groupId&gt;org.apache.maven.plugins&lt;/groupId&gt;     &lt;artifactId&gt;maven-war-plugin&lt;/artifactId&gt;     &lt;executions&gt;         &lt;execution&gt;             &lt;phase&gt;package&lt;/phase&gt;             &lt;configuration&gt;                 &lt;webappDirectory&gt;target/${env}/deploy&lt;/webappDirectory&gt;                 &lt;archiveClasses&gt;true&lt;/archiveClasses&gt;             &lt;/configuration&gt;             &lt;goals&gt;                 &lt;goal&gt;exploded&lt;/goal&gt;             &lt;/goals&gt;         &lt;/execution&gt;     &lt;/executions&gt; &lt;/plugin&gt;   The trouble is, the war file still gets built.  Is there a simple way of having &lt;packaging&gt;war&lt;/packaging&gt; execute the war:exploded goal instead of the war:war goal?  Or is there another simple way to do this? Problem: We have matrix $A\in M_{n-1\times n}(\mathbb Z)$ so that the sum of elements in each row is zero. Prove that $\det(AA^T)=nk^2$, where $k\in \mathbb Z$.  What have I considered so far:   First I thought, since sum of all elements in each row is zero, zero is eigenvalue of $A$ but $A\in M_{(n-1)\times n}(\mathbb Z)$ confused me. I see that $AA^T$ will be $(n-1)$ by $(n-1)$, so I tried calculating $AA^T$ but I failed to see any connection.   I did some research and found  question but we had no mention of it, which makes me believe I am on the wrong track.  Thank you all for your help. Suppose I have a line segment of length $L$.  I now select two points at random along the segment.  What is the expected value of the distance between the two points, and why? I'm new to LaTeX and trying to understand the reason why I'm getting the warning:  `h' float specifier changed to `ht'   I'm not attempting to change the float property of anything - the warning is showing up if I attempt to simply encapsulate some words in quotation marks:  ``highly distributed''   As far as I'm aware, there's no open sections anywhere and the file seems to render properly. So realistically it doesn't matter if the issue gets solved or not; but I'd like to know the reason I'm getting the warning.  Thanks in advance. How to colorize an image as in the example? Evaluating the nested radical $ \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots}}} $. How does one prove the following limit? $$   \lim_{n \to \infty}   \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots \sqrt{1 + (n - 1) \sqrt{1 + n}}}}} = 3. $$ After the events on Miller's planet, it seems to come as a revelation to Brand that     Miller had only landed a couple hours ago, and died a few minutes ago.   The crew of the Endurance, even when they were back at NASA headquarters, had all the information they needed to figure out the time dilation and how long Miller had been on the planet. As such, they must have known that     Miller, even in the best case, would have only had a couple hours worth of data.   Why use all those resources (fuel and time) just for that? Let $X$ be a transitive $G$-set. ($G$ acts on  $X$ transitively.) If $X$ is finite and has at least two elements, show that there is some element $g$ $\in$ G which does not have any fixed points; that is, such that $g$$($$x$$)$ $\ne$ $x$ for all $x$ $\in$ $X$  I am trying to use contradiction, but it is not very clear to me why this is true. What should be done with questions that have been resolved as a comment instead of an answer? I'm trying to go through and provide answers to every unanswered question (search &quot;closed:0 answers:0&quot;), somewhat following the spirit that was blogged about at  .  However, I'm finding this much more difficult than it needs to be, as many of the questions have already been answered without the use of an &quot;answer&quot; - but instead, in the form of a comment by either the OP or another poster. I wouldn't mind reading through and skipping these, as long as there was some appropriate action I could take to at least assist with preventing these questions from being seen again from the same query - either by myself or another SO member.  Is there such an appropriate action?  Should we vote to close?  If so, using what reason? Even if I vote to close, what are the chances that 4 other users will also view the question and also vote to close?  (i.e., will my individual vote even do any good?  Is there any report to show recent close votes, so that others may also review and contribute supporting close votes?) Should we post some sort of an answer - even just something to the effect of &quot;this was already answered by the OP's own comment to his answer (above)&quot;? Would something like this be appropriate to flag for moderator attention, and let the moderators deal with it?  (Or would this be considered too trivial?) Should we rename the question title to something standard that can be used in a search exclusion?  (Maybe &quot;SOLVED:&quot;?)  Basically, how can I help to denote questions as &quot;This question is no longer in active search of an answer - though better answers may still be appreciated.&quot;? I've already reviewed , which was closed as a duplicate of  - but this requires effort on part of a user who may very likely no longer be active. Otherwise, is there something I should be doing to improve my search to filter out these types of questions? I've run across this many times, so I don't consider it to be a rare issue.  However, the last question like this I just ran across, prompting this question, was .  In this case, the question was self-answered by the OP in the form of a comment, this question is the OP's only question, the OP has no other visible activity on SO, and the OP hasn't been seen on SO since he registered and posted the question 5 months ago. Another one I just ran across is .  However, here the &quot;answer&quot; comment was posted by a reputable and recently active SO member.  Should I leave a comment directed at him, asking him to re-post his comment as an answer, or should I post my own answer echoing his comment?  (Or something else?) Yet another is .  Interestingly, the OP actually self-edited the question to include the &quot;SOLVED:&quot; prefix on to the question, instead of posting his resolving comment as an answer (and ideally accepting it) as he was asked to do by another comment.  This brings up another good idea - is there a &quot;standard&quot; we should use to rename the titles on these &quot;resolved but unanswered&quot; posts that could be used as a search exclusion?  (Also added to the above list.) Scrolling has stopped working in chat After cruising through a lot of material online, and answers over here, my understanding of contravariant and covariant vectors are, in a finite-dimensional vector space, suppose we have a vector, which is contravariant to its basis. However, using the metric tensor, we can map this vector to a one form in the dual space, which acts as a vector space for this example. This one form, or covactor, varies similarly to the basis in our original vector space. However, the covectors have a different basis, which is the dual basis, and are expanded in terms of those. Here's where my confusion arises : Many diagrams over the internet give a geometric picture of the scenario, by claiming that contravariant components of an 'arrow' are found by drawing parallel lines along the axes, and checking where they intersect. Co-variant components could be found by sketching the perpendicular lines on these axes. So, these materials are treating covariant and contravariant as different representations of the same arrow, while I'm inclined to believe they are completely different. However, if we let go of rigour, in exchange for a 'more' geometric understanding, I think we are allowed to do this. The diagrams are usually of the form :  However, even if we assume that contravariant and covariant are different representations of the same object, this particular diagram still seems wrong to me. Particularly the locations of $x_1,x_2$. In this diagram, they are located on the span of the original basis. Shouldn't covector components be located on the span of the dual basis ? This diagram suggests to me, that co-variant components have the same basis, as the contravariant ones. Shouldn't the diagram look more like this ?  This second diagram seems to fit the concept better according to me. This is because the dual vector must be 'contravariant' with respect to the 'dual basis'. This means, their components must be found out by sketching parallel lines along the span of the dual basis. These lines intersect the original axis at a right angle, which is to be expected, as dual bases are orthogonal to the original bases. Moreover, this second picture can also show in a much better way, how scaling up of the original basis, scales down the 'contravariant components' and the 'dual basis', which in turn scales up the 'covariant' components.This shows that the components of the dual vector are covariant to the original basis. This is something that is not readily visible in the first diagram. So, am I correct in assuming this second 'geometrical' representation is correct ? I know this doesn't make much sense, because as the mathematicians said, that vectors are completely different from one forms, and one forms should be represented using the number of hyperplanes intercepted by our arrow. However, I've seen most course material refer to it in this way, and frankly it is easier to visualize. However, most of this material used the first picture. Can anyone point out my mistakes, if any, and tell me if the first picture is correct, or the second one ? Or in this case 'less-flawed'. How can I contribute to Ubuntu? I would like to know what kinds of contributions someone can make to Ubuntu. Such as Programmer, User, Proofreader, etc. In the LoTR movies (not sure if the books are the same) but when Gollum lost the ring to Bilbo decades ago in The Hobbit, he appeared to look the same and even move the same in The Fellowship of the Ring. Bilbo on the other hand, seemed to have aged very rapidly when he passed the ring to Frodo. When Frodo met up with Bilbo in Rivendell shortly after his quest began, Bilbo appeared much older and even more much slower.  I realize that Gollum had the ring for much longer than Bilbo, but wouldn't him losing the ring have the same effect as it had on Bilbo? Enable 'on the fly' CRS transformation in QGIS 3? When running sudo apt-get update I get these following errors:  Err archive.ubuntu.com quantal InRelease Err archive.ubuntu.com quantal-updates InRelease Err archive.ubuntu.com quantal-backports InRelease Err archive.ubuntu.com quantal-security InRelease Err archive.ubuntu.com quantal Release.gpg   Unable to connect to archive.ubuntu.com:http: [IP: 91.189.92.177 80] Err archive.ubuntu.com quantal-updates Release.gpg   Unable to connect to archive.ubuntu.com:http: [IP: 91.189.92.177 80] Err archive.ubuntu.com quantal-backports Release.gpg   Unable to connect to archive.ubuntu.com:http: [IP: 91.189.92.177 80] Err archive.ubuntu.com quantal-security Release.gpg   Unable to connect to archive.ubuntu.com:http: [IP: 91.189.92.177 80] Reading package lists... Done W: Failed to fetch archive.ubuntu.com/ubuntu/dists/quantal/InRelease   W: Failed to fetch archive.ubuntu.com/ubuntu/dists/quantal-updates/InRelease   W: Failed to fetch archive.ubuntu.com/ubuntu/dists/quantal-backports/InRelease   W: Failed to fetch archive.ubuntu.com/ubuntu/dists/quantal-security/InRelease   W: Failed to fetch archive.ubuntu.com/ubuntu/dists/quantal/Release.gpg  Unable to connect to archive.ubuntu.com:http: [IP: 91.189.92.177 80] W: Failed to fetch archive.ubuntu.com/ubuntu/dists/quantal-updates/Release.gpg  Unable to connect to archive.ubuntu.com:http: [IP: 91.189.92.177 80] W: Failed to fetch archive.ubuntu.com/ubuntu/dists/quantal-backports/Release.gpg  Unable to connect to archive.ubuntu.com:http: [IP: 91.189.92.177 80] W: Failed to fetch archive.ubuntu.com/ubuntu/dists/quantal-security/Release.gpg  Unable to connect to archive.ubuntu.com:http: [IP: 91.189.92.177 80] W: Some index files failed to download. They have been ignored, or old ones used instead.   I am using Ubuntu 12.04. This is the command I used after installing Ubuntu from window installer. I am sure my laptop is connected to internet. What can I do now? If $M$ is an $R$-module and $I\subseteq\mathrm{Ann}(M)$ an ideal, then $M$ has a structure of $R/I$-module Let $M$ be an $R$-module and let $I$ be an ideal of $R$ such that $I$ is a subset of $\mathrm{Ann}(M)$. Define a product of an element of $R/I$ by an element of $M$ as follows: $$(r+I)\cdot m=rm.$$ (a) Show that this product is well defined. (b) Show that this product, together with the sum of elements of $M$, turns $M$ into an $R/I$-module.  My answer for (a): Let $r+I,s+I \in R/I$ such that $r+I=s+I$. Then $(r+I)m=(s+I)m$, for all $m \in M$. Then $rm=sm \implies r=s$  (since the inverse of $m$ is in $M$), so the product is well defined.  Is my answer to (a) correct?  How can I use Ann(M) to answer (b)? I have to import large Shapefiles (> 1 million records) into PostGIS, and I have been wondering about the best way to do it.    In my question I used the word "hack", instead of tool, on purpose because I think this is not so much a matter of which tool, but which set of steps, or configuration settings to use. So far, I have tried the  plugin (QGIS), the  Postgis tool and the GDAL  tool. You can view my full review on  post. So far, I find all of them really unresponsive, when dealing with a large dataset. I was wondering if someone experienced a similar issue, and if you could share something about the approach. Is it legal/well-defined C++ to call a non-static method that doesn't access members through a null pointer? How to find event listeners on a DOM node when debugging or from the JavaScript code? I've been playing Agar for some time now, and lately I've noticed there was a new XP system in Agar. Since then I've been wondering what variables chance the XP gained. I've tried googling it, but with no luck.  I'm looking for how the XP system works in these game modes:   FFA Teams Experimental