TITLE: Linear Algebra Journal
QUESTION [5 upvotes]: Is there any journal which has significant material on the teaching of linear algebra. I am investigating the most effective way to teach a course on Linear Algebra.
What are the most important things students should learn on a first course?
What are the main questions in this field?

REPLY [1 votes]: Maybe the following isn't helpful, but anyway:
(1) To figure out what the main questions are in linear algebra and which topics to include in a first class, I would simply go find a stack of linear algebra books (hopefully you have access to a nice library) and looks through them and try to identify the topics covered. I would also try to pay attention to how the books get the the same results in different ways.
(2a) Maybe then you could simply just ask yourself how you best learn and try to design your class around that. Students have different preferences, so I usually choose things the way that I like. However, if your preferences is say a very dry, purely abstract treatment, doing no examples, then you might want to ask yourself whether it is likely that your students might appreciate that. As a general principle (that I am sure many people would disagree with) that I have is that I teach best when I figure out what is working for me. I teach the best, when I find the method that I am most comfortable with. If I am trying too hard to be someone that I am not, then that just effects my teaching in a negative way. You might then ask: but shouldn't we evolve and learn new things as teachers? Sure, but if this is the first time teaching a class, you have to start somewhere.
(2b) In stead of going to journals right away, I would also try to contact people who have taught the class in the past and ask for their recommendations.
(2c) You might also be able to find some students who have taken linear algebra and ask them how they think they leaned best in the class.


TITLE: Implication in mathematics - How can A imply B when A is False?
QUESTION [20 upvotes]: I have just started studying implication in mathematics. So, you'll probably have an idea of where my confusion lies right from the get go.
In the truth table, where $A \implies B$ we obtain this result:
A | B |A->B
___________
T | T | T
T | F | F
F | T | T
F | F | T

Now, my confusion here is regarding why $A\implies B$ given A is false, and B true. The others I understand. The first and last one are obvious, the second one implies, to me anyway, that given A implies B, the truth of B rests upon the truth of A, B is false, A is True, which cannot be, thus not B given A is false.
Now... then, why is B true despite the fact that A is false? Or rather, why is the statement B given not A, True. The truth of B is implied from the truth of A, but A is a false statement. So it's very contradictory that this may be, and in fact, if you go through a real life example, you would find that the third statement seems to be false.
Say $A$ is the statement that it is dark outside, and $B$ the statement that the sun is not on our side of the earth. If it is not dark outside, than stating that the sun is on the other side of the planet must be wrong (in my example i'm going to claim that the sun is the only light source).
Side note:
I just had a thought, if A is false, does that mean you can conclude... mmm, whatever? I mean, B is implied via the truth of A, if A is false, B is NOT implied, but then B could be EITHER true or false, I suppose. Is this correct?
However, would that not mean that A does not imply anything when false. Would that not set the 3rd and 4th statement to be undefined?

REPLY [1 votes]: I find it easier to understand implication if I substitute implies with requires.
Consider the example: "the sky is cloudy" and "it is raining".
Notice that "it is raining" requires that "the sky is cloudy". However, "the sky is cloudy" does not require "it is raining".
Using requires instead of implies helps me to distinguish between physical causation (or layman implication) from formal, logical implication.
Furthermore, I even think we should invert the arrow direction, from a --> b to a <-- b, in order to make it easier to use the requires concept.


TITLE: How to solve for $3$ constants with $3$ equations
QUESTION [0 upvotes]: I am working to solve a formula for population, where I have the following model:
$P(t)= M/(1+Ae^{-Mkt})$
I have the first data points for the model:
$(0,2700000), (1,3050000), (2,3450000)$
This gives me $3$ equations:
$2700000=M/(1+A),$
$3050000=M/(1+Ae^{-Mk})$
$3450000=M/(1+Ae^{-2Mk})$
However, I am not sure how to solve for each of the $3$ variables, $M, k$ and $A$?
Any help would be much appreciated.

REPLY [0 votes]: There are many ways to solve the problem. Here is another one : first let $e^{-Mk}=x$ and consider
$$2700000=\frac M{1+A} \tag 1$$
$$3050000=\frac M{1+A x} \tag 2$$
$$3450000=\frac M{1+A x^2} \tag 3$$
$$x=e^{-Mk}\tag 4$$
$$\frac{(1)}{(2)} \implies \frac{54}{61}=\frac{A x+1}{A+1}\implies A=\frac{7}{54-61 x} \tag 5$$
$$\frac{(1)}{(3)} \implies \frac{18}{23}=\frac{A x^2+1}{A+1}\implies x=\frac{144}{161}\implies A=-\frac{1127}{90}\tag 6$$ Now, from $(1)$ $M=-31110000$.
Now, using $(4)$
$$\frac{144}{161}=\exp({31110000 k})\implies k=-\frac{\log \left(\frac{161}{144}\right)}{31110000}$$


TITLE: Algebraically closed concept
QUESTION [0 upvotes]: Suppose $L$ is an extension field of $K$ and $L$ is algebraically closed. Then is it trivially true that $\forall f(x)\in K[x],~f(x)$ splits over $L$? I think the answer is yes since $L$ is algebraically closed means that every polynomial with coefficients in $L$ have a root in $L$ (and hence have all roots in $L$)

REPLY [0 votes]: Yes, the splitting field of $f(x)\in K[x]$ must be contained in $L$ since $L$ is algebraically closed.


TITLE: Sum by twos for functions on $\Bbb{Z}$
QUESTION [0 upvotes]: I have two double sums with the steps $2$ and I do know that one of them is smaller than the other one (due to a complicated argument), but I would like to show it with direct computation.
Let $\sideset{_2}{}\sum$ denote the sum by twos. Then I would like to show that
$$
\sideset{_2}{}\sum_{t=-m}^{m} \sideset{_2}{}\sum_{s=-n}^{n} f(t+s) \geq \sideset{_2}{}\sum_{t=n-m}^{n+m} \sideset{_2}{}\sum_{s=-t}^t f(s).
$$
for  $m,n \in \{0,1,2,3,\ldots\}$ and the function $f:\Bbb{Z}\rightarrow (0,\infty)$.
PS I tried some numbers for $n$ and $m$ and it turned out that all of them give me the equality. But still the inequality is enough for me.

REPLY [1 votes]: Arrange the sum in a careful array:
$$\matrix{f(-m-n)&+f(-m-n+2)&+\cdots&+f(-m+n-2)&+f(-m+n)\cr
          +f(-m-n+2)&+f(-m-n+4)&+\cdots&+f(-m+n)&+f(-m+n+2)\cr
           +\cdots\cr
          +f(m-2)&+f(m)&+\cdots&+f(m+n-4)&+f(m+n-2)\cr
          +f(m)&+f(m+2)&+\cdots&+f(m+n-2)&+f(m+n)\ .\cr}$$
If you now add all the entries in the first column and the last row you get
$$f(-m-n)+f(-m-n+2)+\cdots+f(m-2)+f(m)+f(m+2)+\cdots+f(m+n-2)+f(m+n)$$
which is
$$\sideset{_2}{}\sum_{s=-m-n}^{m+n} f(s)\ .$$
The next column and row give
$$f(-m-n+2)+f(-m-n+4)+\cdots+f(m)+\cdots+f(m+n-4)+f(m+n-2)$$
which is
$$\sideset{_2}{}\sum_{s=-m-n+2}^{m+n-2} f(s)\ ,$$
and so on.  You may have to break up into cases depending on which of $m,n$ is bigger, but if for example $n\ge m\ge0$ then the last sum will involve the leftover items from the top row only and will be
$$f(m-n)+\cdots+f(-m+n)=\sideset{_2}{}\sum_{s=m-n}^{-m+n} f(s)\ .$$
These partial sums (reading them in the reverse order from how we found them) are
$$\sideset{_2}{}\sum_{s=-t}^{t} f(s)$$
for
$$t=n-m,\ldots,n+m-2,n+m$$
and so we get exactly your RHS.  In conclusion: as you conjectured, the two double sums are equal.


TITLE: regular singularities
QUESTION [0 upvotes]: Hi friends,
Let me ask you about connexions having regular singularities. So imagine $X$ is some smooth algebraic variety over a subfield $k \subset \mathbb{C}$ and you have a locally free $\mathcal{O}_X$-module $\mathcal{E}$ with a flat connection $\nabla: \mathcal{E} \to \mathcal{E} \otimes \Omega^1$. This defines a complex $DR(\mathcal{E}, \nabla)$ and you can consider
$H^n(X, (\mathcal{E}, \nabla)):=\mathbb{H}^n(X, DR(\mathcal{E}, \nabla)).$
On the other hand, there is an associated analytic connection $(\mathcal{E}^{an}, \nabla^{an})$ on $X^{an}$ and the corresponding cohomology
$H^n(X^{an}, (\mathcal{E}^{an}, \nabla^{an})).$
There is of course a morphism (*)
$H^n(X,(\mathcal{E}, \nabla)) \otimes_k \mathbb{C} \to H^n(X^{an}, (\mathcal{E}^{an}, \nabla^{an}))$.
When $(\mathcal{E}, \nabla)$ is the trivial connection $(\mathcal{O}_X, d)$, then you just get algebraic and analytic de Rham cohomology, which are isomorphic by Grothendieck's theorem. In general, easy examples show that the above morphism is not an isomorphism.
The condition needed to guarantee that it (*) is an isomorphism is that the connection has regular singularities, meaning that there exists a logarithmic extension
$$
\nabla^{ext}: \mathcal{E}^{ext} \longrightarrow \mathcal{E}^{ext} \otimes \Omega^1(\log D)
$$
to a good compactification $\bar{X}$ including $X$ as the complement of a normal crossings divisor.
Then Deligne proved that if $(\mathcal{E}, \nabla)$ has regular singularities, there is an isomorphism
$H^n(X,(\mathcal{E}, \nabla)) \otimes_k \mathbb{C} \stackrel{\sim}{\longrightarrow} H^n(X^{an}, (\mathcal{E}^{an}, \nabla^{an}))$.
I'm trying to understand what does it mean for a very simple connection, constructed on the trivial bundle on $\mathbb{G}_{m, k}$. To give a connection on $\mathcal{O}$ is the same as giving a differential $1$-form, so assume
$\nabla(1)=\alpha \frac{dt}{t}$
for some element $\alpha \in k^\times$. I would say that this connection has regular singularities because $\frac{dt}{t}$ has simple poles at $0$ and $\infty$, so $\mathcal{O}$ of $\mathbb{P}^1$ with $\nabla(1)=\alpha \frac{dt}{t} \in \Omega^1(\log (0+\infty))$ is a logarithmic extension.
My question is: how does one see that this two cohomologies are isomorphic in this particular case? And what is a description of the local system $\ker(\nabla^{an})$ underlying the connection?
Thanks!

REPLY [1 votes]: We can see what $\ker(\nabla^\mathrm{an})$ will be by the following reasoning (which can be made a bit more precise):
$\nabla(f)=0\Leftrightarrow \frac{df}{f}=-\alpha\frac{dt}{t}\Leftrightarrow \log(f)=-\alpha\log(t)+c\Leftrightarrow f=c_0t^{-\alpha}$. So local sections of $\ker(\nabla^\mathrm{an})$ will then just be multiples of some fixed branch of the function $t\mapsto t^{-\alpha}$.
Global sections of this will exist iff $\alpha\in\mathbb{Z}$, and so in this case we can see that $H^0(\mathbb{G}_m^\mathrm{an},(\mathcal{O}^\mathrm{an},\nabla^\mathrm{an}))$ will be zero if $\alpha\notin\mathbb{Z} $ and $\mathbb{C}\cdot t^{-\alpha}$ otherwise.
On the other hand, algebraic de Rham cohomology will just be the cohomology of the complex of $k[t,t^{-1}]$-modules
$0\rightarrow k[t,t^{-1}]\overset{f\mapsto f'+\frac{f\alpha}{t}}{\rightarrow} k[t,t^{-1}]\rightarrow0$
Thus elements of $H^0(\mathbb{G}_{m,k},(\mathcal{O},\nabla))$ will be (multiples of) global algebraic branches of $t\mapsto t^{-\alpha}$. Again, if $\alpha\in\mathbb{Z}$, then this $H^0$ is just $k\cdot t^{-\alpha}$ and if $\alpha\notin\mathbb{Z}$ it will be zero.
I'm not sure how to see 'directly' that the two $H^1$'s will be the same though.


TITLE: Problems about the convex hull.
QUESTION [2 upvotes]: I'm stuck in two problems concerning about convex hull.

Let $A,B,C  \not= \emptyset$, compact sets in $\mathbb{R^n}$. Show that if $A+B=A+C$ then $\text{conv}(B)=\text{conv}(C)$
Let $B\not= \emptyset$ in $\mathbb{R^n}$. Show that $n \text{conv}(B)+B=(n+1)\text{conv}(B).$ Also show that this is not true if we take some $m<n$.

I would appreciate some tip.

REPLY [3 votes]: This is too long for a comment, so I'm posting it as a (partial) answer. In general, it is true that $\text{conv}(A+B)=\text{conv}(A)+\text{conv}(B).$ And this is enough to prove the first part:
$t(A+B)+(1-t)(A+B) = (tA+(1-t)A) + (tB+(1-t)B) = A+B$, so the sum $A+B$ is convex, and since it also contains $A+B$, it must contain the convex hull of $A+B.$
On the other hand, if $x\in \text{conv}(A+B),$ then it is a finite sum $\sum t_i(a_i+b_i)$ for some $a_i\in A;\ b_i\in B;\ t>0;\  \sum t_i=1.$
Then, $\sum t_i(a_i+b_i)=\sum t_ia_i+\sum t_ib_i$ so $x\in \text{conv}(A)+\text{conv}(B).$


TITLE: Existence of a  symmetric matrix.
QUESTION [0 upvotes]: In the real field. Given a diagonal matrix $D$ and a symmetric matrix $A$. For every skew symmetric matrix $S$, is there always a symmetric matrix $H$ such that $-\operatorname{trace}(DSAS)=\operatorname{trace}(DHAH)$?
If $A$ is also diagonal, this can be easily seen true.

REPLY [4 votes]: Maybe I'm missing something, but let $f(H)=\mathrm{tr}(DHAH)$, then
$$f(\alpha H)=\alpha^2f(H)$$
for every real $\alpha$. Obviously $f(0)=0$. So the problem boils down to find an $H$ such that $f(H)>0$ and another one such that $f(H)<0$.
Now, denoting by $h_k$ the $k-$th column of $H$,
$$(HAH)_{ij}=(h_i)^tAh_j$$
Therefore $f(H)=\sum d_i (h_i)^tAh_i$, if $D=\mathrm{diag}(d_1,\ldots, d_n)$.
If, wlog, $d_1>0$ and $d_2<0$, then we are obviously done.
If every $d_i$ has the same sign, but $A$ has a positive and a negative eigenvalue, then again we are done.
If every $d_i$ is wlog positive (or null) and $A$ is semi-definite, then $f(H)\geq0$ for every $H$ symmetric and $f(S)\leq 0$ for every $S$ skew-symmetric, so, given $H$ such that $f(H)\neq0$, we can always find $\alpha$ such that $f(\alpha H)=-f(S)$.


TITLE: For a (left) ideal $I$, how can $R \cdot I$ be a proper subset?
QUESTION [1 upvotes]: I have the following definition for a left ideal: An additive subgroup $I$ of a ring $R$ is called left ideal, if $R \cdot I \subseteq I$ where $R \cdot I = \{ r \cdot i \,\,\,|\,\,\, r \in R, i \in I\}$.
My question is: how can $R \cdot I$ be a proper subset of $I$? My first observation is that if this is indeed the case, $R$ isn't allowed to have a neutral element for multiplication, because else $R \cdot I \supseteq 1 \cdot I = I \Rightarrow R=I$.
Is there an easy example?

REPLY [2 votes]: Take $R = 2\mathbb Z, I = 8 \mathbb Z.$ Then $8 \notin R \cdot I.$


TITLE: Growth of smallest closed geodesic in congruence subgroups?
QUESTION [11 upvotes]: Let $\Gamma$ be one of the classical congruence subgroups $\Gamma_0(N)$, $\Gamma_1(N)$ and $\Gamma(N)$ of $SL(2, \mathbb{Z})$.
How does the lower bound for the length of primitive geodesics on $\Gamma \backslash \mathbb{H}$ depending on $N \rightarrow \infty$?
Any suggestions?

REPLY [11 votes]: For a hyperbolic element $A\in SL(2,\mathbb{Z})$, we have the length of the closed geodesic is given by $\ln[(tr^2(A)-2+\sqrt{tr^4(A)-4tr^2(A)})/4]$, and this is monotonic in $|tr(A)|$ for $|tr(A)|>2$. For $A\in \Gamma(N),\Gamma_1(N)$, we have $tr(A)\equiv 2 (\mod N)$, and $tr(A)\neq \pm 2$, so the smallest that $|tr(A)|$ can be is when $|tr(A)|=N-2$ (I suppose for $N>4$).  This gives a lower bound on the shortest geodesic for $\Gamma_1(N)$. This is realized by the matrix
$$A=\left[\begin{matrix} 1-N & 1 \\ -N & 1 \end{matrix}\right]$$
For $\Gamma(N)$, one can obtain a better lower bound. Consider the matrix $$A=\left[\begin{matrix} 1+aN & bN \\ cN & 1+dN \end{matrix}\right]$$ with $\det(A)=1$. Then we have $(1+aN)(1+dN)-bcN^2=1$, so $a+d+(ad-bc)N=0$. This implies that $a+d\equiv 0(\mod N)$, so $tr(A)=2+(a+d)N \equiv 2 (\mod N^2)$. Thus, we get a lower bound of $|tr(A)|\geq N^2-2$. This is realized by the matrix
$$A=\left[\begin{matrix} 1-N^2 & N \\ -N & 1 \end{matrix}\right]$$
Edit: (I'm modifying the answer to address Vitali's question in the comments below).
For a matrix $A\in \Gamma_0(N)$, we have
$$A=\left[\begin{matrix} a & b \\ cN & d \end{matrix}\right]$$ with $\det(A)=1$.
 Then $ad-bcN=1$ implies $ad \equiv 1(\mod N)$. We want to minimize $tr(A)=a+d$ subject to the constraint $ad\equiv 1(\mod N)$. Conversely, if $ad=1+kN$ for some $k$, then the matrix
$$A=\left[\begin{matrix} a & k \\ N & d \end{matrix}\right]\in \Gamma_0(N)$$
has trace $a+d$. So the minimal trace of a hyperbolic element in $\Gamma_0(N)$ is given by $\min \{ a+d >2 | ad\equiv 1 (\mod N)\}$.
Let's reformulate this problem.
$ad\equiv 1(\mod N)$ is equivalent to the characteristic polynomial $\lambda^2-tr(A)\lambda+1\equiv(\lambda-a)(\lambda-d) (\mod N)$, i.e. the characteristic polynomial of $A$ reduces $(\mod N)$. So we want to  minimize
$\min \{ t > 2 | \lambda^2-t\lambda+1 \equiv 0 (\mod N), some \lambda \}$.
If $t$ is even, then we complete the square to get $(\lambda-t/2)^2 \equiv t^2/4-1 (\mod N)$, that is $t^2/4-1$ is a quadratic residue $(\mod N)$. If $t$ is odd, then $N$ must be odd if $\lambda^2-t\lambda+1\equiv 0 (\mod N)$, so multiplying by $4$, this is equivalent to $(2\lambda-t)^2\equiv t^2-4 (\mod N)$.
Thus, the minimal trace is given by
$\min \{ t>2 | t^2-4$ is a quadratic residue $(\mod N)$, $N$ odd,  or $t^2/4-1$ is a quadratic residue $(\mod N)$, $N$ even $\}$.
Thus, since there are infinitely many $N$ such that $3^2-4=5$ is a quadratic residue $(\mod N)$ (e.g. the sequence $N=a^2-3a+1$), we have that the systole does not approach $\infty$.
Also, the systoles are unbounded from above. To see this, note that if $j$ is not a quadratic residue $(\mod N)$, then it is not a quadratic residue $(\mod kN)$ for any $k$. For $t>2$, choose $n(t)$ such that $t^2-4$ is not a quadratic residue $(\mod n(t))$. Then the number $N(T)=n(3)n(4)\cdots n(T)$ has the property that the minimal trace of $\Gamma_0(N(T))$ is bigger than $T$. In particular, the systole of $\Gamma_0(N!)$ $\to \infty$.

REPLY [3 votes]: For more general results and extensive references, see Katz, Schaps, Vishne, Logarithmic growth of systole of arithmetic riemann surfaces along congruence subgroups (JDG, 2007)


TITLE: Can all quadric surfaces be obtained by cutting a 4-dimensional cone?
QUESTION [5 upvotes]: In my high school multivariable calculus class, we recently learned of quadric surfaces. Since they appeared to be a generalization of conic sections to 3 dimensions, I wondered if they could be generated by finding the intersection of a 3-dimensional plane with a 4-dimensional cone similarly to how a conic section is the intersection of a 2-dimensional plane with a 3-dimensional cone. What follows is my attempt at experimenting with this idea. Although I was able to obtain some interesting results, I was not entirely successful in my endeavour which is why I am posting here.
I was able to deduce that a fourth dimensional equivalent of a circular cone is:
$$x^2+y^2+z^2=w^2$$
With this, I was able to get a few quadric surfaces with minor effort by simply plugging in a constant for one of the variables. For example, by setting $w=3$, I got the equation of a sphere:
$$x^2+y^2+z^2=9$$
And by setting $z=3$, I was able to derive the equation of a hyperboloid of two sheets:
$$x^2+y^2+9=w^2$$
$$w^2-x^2-y^2=9$$
I wanted to see if I could get the rest of the quadric surfaces by picking less mundane planes than just those aligned with the axes. So then I attempted to define a 3-dimensional plane in a 4-dimensional space with a point and a normal. I remembered that one definition for a plane is:
$$\vec{n} \cdot (\vec{P_0} - \vec{P})=0$$
Where $\vec{n}$ is the normal of the plane, $\vec{P_0}$ is the initial point, and $\vec{P}$ is any arbitrary point that lies on the plane. Armed with this knowledge, I plugged in  $\vec{n}=\langle A,B,C,D \rangle$ and $\vec{P}=\langle x,y,z,w \rangle$ to arrive at:
$$Ax+By+Cz+Dw=\vec{n} \cdot \vec{P_0}$$
Since the cone should have rotational symmetry (admittedly I am not sure what this means in four dimensions so this could be a faulty assumption), I figured that any point would yield the same results as long as it does not lie on the plane $w=0$. With this knowledge, I picked the point $(1,0,0,1)$ because it seemed to be easy to work with.
Then, I tried various normals that I thought would yield interesting results. With this method, I was able to generate most quadric surfaces (including elliptic paraboloids). However, no matter what I tried, I was not able to generate a hyperbolic paraboloid or a hyperboloid of one sheet. Am I not picking the right normals for this? Or is my reasoning flawed? Could it be the case that it is actually not possible to generate these quadric surfaces by cutting a 4-dimensional cone with a plane?

REPLY [4 votes]: In dimension $3$, then, there's essentially only one cone, namely, the zero locus of a nondegenerate, indefinite quadratic form $Q$. Sylvester's Law of Inertia says that by making choosing suitable linear coordinates (and possibly replacing $Q$ by its negative, which doesn't change the zero locus), we can always put the cone in standard form: $$x^2 + y^2 - z^2 = 0 .$$
In dimension $4$, however, there are two inequivalent cones, with standard forms $$w^2 + x^2 + y^2 - z^2 = 0 \qquad \textrm{and} \qquad w^2 + x^2 - y^2 - z^2 = 0 .$$ (The corresponding quadratic forms respectively have signature $(3, 1)$ and $(2, 2)$, sometimes called Lorentzian and neutral, respectively.)
We can now recover the two missing quadric surfaces as hyperplane sections of the neutral cone, $C$:

Intersecting a neutral cone $C$ with any hyperplane with spacelike normal vector ${\bf n}$ (i.e., $Q({\bf n}) > 0$) and not passing through the origin gives the a hyperboloid of one sheet. For example, taking the standard form and the hyperplane $\{z = 1\}$, and using $(w, x, y)$ as coordinates on the hyperplane gives the standard form $$w^2 + x^2 - y^2 = 1 .$$

Instead intersecting $C$ with a suitable hyperplane parallel to any $2$-plane contained in the cone (and again not passing through the origin) gives a hyperbolic paraboloid. For example, taking the standard form and the hyperplane $\{w = z + a\}$, $a \neq 0$, gives the surface $w = z + a, x^2 + 2 a z + a^2 = y^2$. If we use $(x, y, z)$ as coordinates on the hyperplane, applying the affine change $x, y, z' := -(2 a z + a^2)$ of coordinates there puts the equation in the standard form $$z' = x^2 - y^2$$ of a hyperbolic paraboloid.


TITLE: Is SO(4) a subgroup of SU(3)?
QUESTION [33 upvotes]: $\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}$I want to write a $3 \times 3$ complex-matrix representation of $\SO(4)$, for example, we know that $\SO(5)$ is a subgroup of $\SU(4)$, so we can write a $4 \times 4$ complex-matrix representation of $\SO(5)$. There is a paper which gives this representation: Mapping two-qubit operators onto projective geometries. This paper mentions which matrices form the $\SO(5)$ group, and Manipulating two-spin coherences and qubit pairs contains the matrices explicitly (both papers are by A.R.P. Rau).
I want to know if $\SO(4)$ is also a subgroup of $\SU(3)$, and if so, can we find a $3 \times 3$ complex-matrix representation of $\SO(4)$?

REPLY [12 votes]: In addition to all the completely correct answers to this question, it's probably worth spelling out some of the misconceptions that are common to physicists who work in this area and giving the physics answer.
The main thing that arises all the time is that physicists often conflate representations of a group and of its Lie algebra. The reason one can get away with this is that really in physics one deals with projective representations which are the same (in most common cases) as representations of the universal cover of the group. Thus, distinctions like the difference between $Spin(4) \simeq SU(2) \times SU(2)$ and $SO(4) \simeq \left(SU(2) \times SU(2)\right) / \mathbb{Z}_2$ are lost.
The other slightly odd (and not so common) misconception that seems to be here is that you need a group to be a subgroup of $U(n)$ to have a complex $n$ dimensional representation. A map into $U(n)$ is more than enough for obvious reasons.
Taking these into account, the original statement in the question becomes that you have a map from $Spin(5) \simeq Sp(4) \to SU(4)$ giving a four dimensional representation. And the answer to the physics version of the question is "yes-ish", because there is a map from $Spin(4) \to SU(2)$ given by projection onto one of the two $SU(2)$ components (say) and then you can take the $\mathbb{Z}_2$ quotient to get a map from $SO(4) \to SO(3)$.
Or, to put it another way, there is a perfectly cromulent, but not faithful, three dimensional representation of $SO(4)$.


TITLE: Rotation of angular momentum eigenfunctions?
QUESTION [2 upvotes]: I am struggling to understand this apparently obvious example in my group theory notes:

Where do the $e^{i\phi} $ and $e^{-i\phi} $ factors come from?
I know that the $m_l$ = -1,0 and +1 angular momentum eigenstates have the $\phi$ dependence incorporated into a complex exponential factor of the very same form... But should the $\phi$ there be a variable? As opposed to a fixed rotation angle?

REPLY [4 votes]: An angular momentum eigenstate can be rotated using,
\begin{equation}
\left| J , m  \right\rangle  \rightarrow  e ^{ i {\vec S}   \cdot {\vec \theta} } \left| J , m  \right\rangle
\end{equation}
where $ {\vec S} $ is the $2J+1$ dimensional Pauli matrices. For spin $ 1/2 $ for example, $ {\vec S}  $ are just the ordinary Pauli matrices, $ \frac{1}{2} {\vec\sigma} $. The vector $ \theta $ parameterizes the rotation. Depending on the dimensionality of the value of $ J $ the dimensionality of $ {\vec S} $ is different.
For spin $1$, the $S$ matrices are:
\begin{equation}
S _x = \frac{1}{\sqrt{2}} \left( \begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 1 \\
0 & 1 & 0
\end{array} \right) , \quad S _y = \frac{1}{\sqrt{2}} \left( \begin{array}{ccc}
0 & - i  & 0 \\
i & 0 & - i  \\
0 & i  & 0
\end{array} \right) , \quad S _z = \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -1
\end{array} \right)
\end{equation}
Under a rotation only in the $z$ direction ($ {\vec \theta} = (0,0, \phi ) $) we only need $ S _z $ and we have,
\begin{equation}
\left| 1 , m _j \right\rangle  \rightarrow  e ^{ i S _z \phi  } \left| 1 , m  \right\rangle
\end{equation}
Since $ S _z $ is diagonal its trivial to exponentiate,
\begin{equation}
e ^{i S _z \phi  } = \left( \begin{array}{ccc}
e ^{ i \phi }  & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & e ^{ - i \phi }
\end{array} \right)
\end{equation}
giving the transformation you want.