How is the trace of the adjoint of Lie algebra elements defined? Consider an element $X\in\mathfrak g$ of some Lie algebra $\mathfrak g$. I understand that $\mathfrak g$ can be represented via its action on other elements of the same algebra, as $\operatorname{ad}(X)Y\equiv[X,Y]$, so that $\operatorname{ad}(X)\in\operatorname{GL}(\mathfrak g)$.  One often considers things such as the trace of these objects (e.g. when considering the Killing form), or more generally the "matrix elements" of operators such as $\operatorname{ad}(X)$.  However, I usually don't see any direct mention of the inner product with respect to which these things are defined. To properly define what something such as $\operatorname{ad}(X)_{ij}$ is, don't I need to be able to define uniquely the coefficient of the generator $X_i$ in the decomposition of $[X,X_j]$ (denoting with $\{X_i\}$ the elements of some basis for the algebra)?  Is there a canonical choice of such inner product? And on a similar note, do we just usually assume that the basis of the Lie algebra is orthonormal (or at least orthogonal) with respect to this inner product?
Coordinate-Free Definition of Trace. $\DeclareMathOperator{\tr}{trace}$ I am reading the  on the trace operator. The section titled Coordinate-Free Definition defines the trace as follows.  Let $V$ be a finite dimensional vector space over a field $F$ and define a bilinear map $f:V\times V^*\to F$ as $f(v, \omega)=\omega(v)$ for all $(v, \omega)\in V\times V^*$. This maps induces a unique linear map $\tr:V\otimes V^*\to F$. Since $\text{End}(V)$ has a canonical isomorphism with $V\otimes V^*$, we have now a notion of trace of a linear operator on $V$.  The Question: The second paragraph of the section in the article says that  This also clarifies why $\tr(AB)=\tr(BA)$.  I can't see how $\tr(AB)=\tr(BA)$ follows from this definition at all. Can somebody give me a hint?
fake proof of $\forall a. \forall b. a = b \to 1 = 0$ I saw a less formal version of this fake proof that claimed to prove $2=1$ but because it assumed $a=b$ from the start I knew why it was wrong. It does seem however that the proof can be used to prove an certain implication I find very odd however. Clearly $\forall a. \forall b. a = b \to 0 = 1$ is false because I could pick $a$ and $b$ to be $1$ and then I would have derived 0 = 1 with no other conditions. So there must be something wrong with this proof that I am not seeing.  proof of $\forall a. \forall b. a = b \to 0 = 1$:  $a = b \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (assume)\\ a+a=a+b  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (x=y \to f(x) = f(y), f(x)=a+x)\\ 2a=a+b \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  (a+a=2a)\\ 2a-2b=a+b-2b \,\,\, (x=y \to f(x) = f(y), f(x)=x-2b)\\ 2(a-b)=a+b-2b \,\, (distributive)\\ 2(a-b)=a-b \,\,\,\,\,\,\,\,\,\,\,\,\,\, (-b=b-2b)\\ 2=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (x=y \to f(x) = f(y), f(x) =x/(a-b))\\ 1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (x=y \to f(x) = f(y), f(x) =x-1) $  now that I have "proven" $\forall a. \forall b. a = b \to 0 = 1$ I can now pick a and b to be 1.  $ 1 = 1 \to 1=0 \,\,\,\,\,\, (a=1, b=1, above)\\ 1 = 1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\text{reflexivity})\\ 1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\text{modus ponens}) \\ $  So now it seems that I have managed to derive $1=0$. Clearly I messed up given that no one has been able to show $ZFC$ inconsistent and ostensibly this fits well within the most studied aspects of $ZFC$. Either that or a random guy that makes meme's online found the most profound result in mathematics ever.  So where have I missed the mistake? Why align and equation environment do not tolerate empty lines? How to recognise/thank great cabin crew? Those of us who fly a lot will know that there is a bit of a variety in cabin crew. Some (sadly more common on certain airlines) do the very bare minimum to ensure a safe flight, but nothing more. Many (especially on airlines noted for good customer service) provide a good service.  Every so often though, you come across a flight crew who go the extra mile. These are the ones always friendly, always helpful, always thinking ahead and solving the problem before it has happened. One moment they're serving you your drink, the next reassuring a little old lady about the flight, the next magicing a colouring book to turn an impending toddler crisis into a happy small child, then answering queries on arrivals and immigration for a first time travaller, then back to offer you a top up of what you're drinking (which they've remembered exactly). Not very often found, but very much appreciated when you happen across them on your travels!  In the case when you do luck out and find yourself on a flight with a great cabin crew (who go well beyond what you normally experience as "good" on that airline/kind of airline), is there a standard/common way to recognise them at the end of the flight? Not just to thank them then and there, but also to let the airline know that this person/group are what the rest of their staff should be aspiring to. Is there any standard way, or does it vary entirely by carrier? Express partial derivatives of second order (and the Laplacian) in polar coordinates $z=f(x,y)$ where $x=rcosθ$ and $y=rsinθ$  Find $ \frac{\partial z}{\partial x}$ and $ \frac{\partial^2 z}{\partial x^2}$  I'm having big troubles with using chain rule, in particularly the second derivative. Spent 2 hours on this already. What I have...  $ \frac{\partial z}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial z}{\partial r}+\frac{\partial z}{\partial θ}\frac{\partial θ}{\partial x}$. So I'm guessing that $\frac{\partial z}{\partial r}$ and $\frac{\partial z}{\partial θ}$ are unkown? And I will need to use them to differentiate again. Any help is really welcome. This is simply only part of a bigger question.   Full Question: How to set the default program for opening files without an extension in Windows? How to set the default program for opening files without an extension in Windows? 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The code is the following:  @Test public void rowSelection() throws Exception {     SeleniumHelper helper = new SeleniumHelper();     action = new SeleniumActionHelper(driver);      helper.login();      String testUrl = navigateToUrl("option/listOptions.xhtml");     driver.get(testUrl);      WebElement table = findElementById("tableSection:dataTableWrapped_data");     List&lt;WebElement&gt; allRows = table.findElements(By.tagName("tr"));     for (WebElement row : allRows)         {             List&lt;WebElement&gt; cells = row.findElements(By.tagName("td"));             for (WebElement cell : cells)             {                 WebElement listName = cell.findElement(By.xpath("./* [text()='body_build']"));                 listName.click();             }         } }   I have put a Thread.sleep(2000) in front and after the listName.click() action, but neither has helped. Any help would be greatly appreciated.