I've seen a definition of projective representation as this: Given a group $G$, a projective representation can be defined with a map
$$\pi:G\rightarrow GL(\mathbb{V})$$
from the group to the linear operators in some $\mathbb{k}$-valued vector space $\mathbb{V}$ and a function
$$\lambda: G\times G\rightarrow \mathbb{k}^*$$
that takes two elements in $G$ and gives back multiplication by a $\mathbb{k}$ number. The product in this representation is
$$\pi(g_1)\pi(g_2)=\lambda(g_1,g_2)\pi(g_1\cdot g_2)$$
My Questions
First part of my question is: Do projective representations satisfy the other rules of representations? That is:
- The map $\pi$ takes the neutral element $g_e\in G$ to the neutral element in $GL(\mathbb{V})$
$$\pi(g_e)=\mathbb{I}$$
- The representation of the inverse is the inverse of the representation
$$\pi(g^{-1})=\big[\pi(g)\big]^{-1}$$
The reason I ask these questions is because I tried manipulating the product rule and I get confusing results. For example, if I use the product rule for an element $g\in G$ and the neutral element $g_e\in G$ I get
\begin{equation} \pi(g)\pi(g_e)=\lambda(g,g_e)\pi(g) \end{equation}
which makes me think that multiplication by the identity is given by
$$\pi(g_e)=\lambda(g,g_e)$$
But this equation confuses me as well. Am I supposed to understand it as $\pi$ being a multivalued map that takes $g_e$ to $\lambda(g,g_e)$ with $g$ all elements in $G$? Or am I supposed to understand that the identity is represented by a different scalar $\lambda(g,g_e)$ depending on what we multiply it with?
On a similar line, I've seen the projective representation as the equivalence class of all operators in $GL(\mathbb{V}$ that differ in a multiplicative factor. But that seems to contradict the product rule I wrote earlier, where the multiplicative factor seems to be determined by the elements we are multiplying. So is the projective representation a set of equivalence classes as well?
Finally, I understand that by multiplying three elements $g_1$, $g_2$ and $g_3$ and associating them indifferent ways we get a constraint on $\lambda$
$$\lambda(g_1,g_2)\lambda(g_1\cdot g_2,g_3)=\lambda(g_1,g_2 \cdot g_3 )\lambda(g_2,g_3)$$
However, if we pick $g_1=g$, $g_2=g^{-1}$ and $g_3=g_e$ we get
$$\lambda(g,g^{-1})\lambda(g_e,g_e)=\lambda(g,g^{-1})\lambda(g^{-1},g_e)$$
which gives the weird equation $\lambda(g_e,g_e)=\lambda(g^{-1},g_e)\ \ \forall g\in G$. I don't even know what to make of this equation.
I'm a theoretical physicist so I get a bit of basic group theory but not much else! Thanks!
I have rewritten my answer.
Let $V$ be a vector space on a field $k$. Let us recall that if $V$ is a vector space, $\mathbb{P}(V)$ is the set of equivalence classes of vectors up to the multiplication by a nonzero constant. The equivalence classes are just lines (without the origin). Note that in a line, there is no preferred vector: any nonzero vector contained in that line is enough to reconstruct the line, and all of them are equally nice (if you are allowed to use the concept of a norm, you can say that norm $1$ vectors are better than the others, but you still have to choose between at least to norm $1$ vectors!). If $A \in GL(V)$, then $A$ maps two collinear of vectors to collinear vectors; so $A$ induces a bijection of $\mathbb{P}(V)$. We denote this induced bijection $\mathcal{P}(A)$ (this is not a universal notation, I just made it up). The set of all such bijections is denoted by $PGL(V)$, is a group, and $\mathcal{P}$ is a surjective morphism; each of the fibers are "lines" in $GL(V)$, that is, equivalence classes with respect to the relation "being equal up to a nonzero multiplicative constant". For example, let us consider $k = \mathbb{R}$, $V = \mathbb{R}^2$. Then $GL(V)$ is (isomorphic to) the group of invertible $2\times 2$ matrices, and an element of $PGL(V)$ is the set $\{\begin{pmatrix}a &0&0&a\\\end{pmatrix} \ \vert \ a \in \mathbb{R}^*\}$. In this case, this set has a preferred element, the identity matrix. However, does $\{\begin{pmatrix}2a&a\\a&a\\\end{pmatrix} \ \vert \ a \in \mathbb{R}^*\}$ have a preferred element? In general, there is no natural way to choose an element of the equivalence class to represent the equivalence class (be careful, we use represent in the informal sense, not the mathematical one). So, in general, one cannot describe more precisely the elements of $PGL(V)$ than being equivalence classes. For all these reasons, I, personally, like $V$ better than $\mathbb{P}(V)$: I like vector spaces. I find $V$ easier than $\mathbb{P}(V)$. And I like $GL(V)$ more than $PGL(V)$!
Let $G$ be a group. A map $\lambda : G \times G \rightarrow k^*$ is said to be a cocycle if it satisfies: $\forall g,h,k \in G^3,\quad \lambda(g,h)\lambda (gh,k) = \lambda(g,hk)\lambda(h,k)$. It is said to be a coboundary if there exists a map $\mu : G\rightarrow k^*$ such that $\forall g,h \in G^2,\quad \lambda(g,h) = \frac{\mu(g)\mu(h)}{\mu(gh)}$. It is an important exercice to check that any coboundary is a cocycle. We can already notice a property of cocycles: let $\lambda$ be a cocycle. Then, for all $g$, $\lambda(e,g) = \lambda(g^{-1}, e)$ (exercise). Moreover, for all $g$, $\lambda(g,e) = \lambda(e,g) \lambda(e,e)$. Indeed, let $g \in G$. We have $\lambda(e,g)\lambda(g, g^{-1}) = \lambda(e, g^{-1}g) \lambda(g,g^{-1})$, and therefore $\lambda(e,g) = \lambda(e,e)$.
A representation is a morphism $G \rightarrow GL(V)$, and a projective representation is a morphism $G \rightarrow PGL(V)$. Since I like $GL(V)$ more than $PGL(V)$, I also like representations more than projective representations. Let $\pi$ be a representation. Then $\rho := \mathcal{P} \circ \pi$ is a projective representation. In this case, we say that $\pi$ gives rise to $\rho$. Now, let $\rho$ be a projective representation. Since I like representations more, how can I find a representation $\pi$ that gives rise to $\rho$? Does there even exist one? Is there something that could play the role of a projective representation that is closer to what I like?
First of all, one can always (with some choice, but it shouldn't be very important here) find sections of $\mathcal{P}$, that is, maps $s : PGL(V) \rightarrow GL(V)$ such that $\mathcal{P}\circ s = Id_{PGL(V)}$. Let $s$ be any section, and let $\rho$ be a projective representation. Let us define $\pi := s \circ \rho$. Then $\mathcal{P} \circ \pi = \rho$, we are halfway there! However, is $\pi$ a representation? If it not a representation, the only way we have to make it a representation is to find a suitable map $c : G \rightarrow k^*$, define $\pi'(g) := c(g) \pi(g)$ for all $g$, hoping that $\pi'$ is turned into a representation. Let us find out.
First of all, let us denote by $e$ the neutral element of $G$. Do we have $\pi(e) = Id_V$? In general, no! It depends on $s$. We only know that there is $c \in k^*$ such that $\pi(e) = c Id_V$. This is manageable: we can replace every $\pi(g)$ by $\pi'(g) := c^{-1}\pi(g)$ and, there you go, $\pi'(e) = Id_V$. This problem is easy to solve.
Second of all, do we have, for all $g$, $\pi(g)^{-1} = \pi(g^{-1})$? In general, no! It depends on $s$. We only know that, for all $g$, there exists $c(g)$ such that $\pi(g)^{-1} = c(g) \pi(g^{-1})$. Computing $\pi(g^{-1})^{-1}$ in two different ways gives us: $\forall g,\quad c(g) = c(g^{-1})$. This is less manageable: let $f$ be a square root of $c$, i.e. a map such that $\forall g, \quad f(g)^2 = c(g)$. Then we can replace every $\pi(g)$ by $\pi'(g) := f(g)\pi(g)$. Indeed, we then have $\pi'(g)^{-1} = f(g)^{-1} \pi(g)^{-1} = f(g)^{-1} c(g) \pi(g^{-1}) = f(g) \pi(g^{-1}) = f(g^{-1}) \pi(g^{-1}) = \pi'(g^{-1})$. I say less manageable, because if we impose continuity restrictions, such a map $f$ might not exist.
Finally, do we have, for all $g,h$, $\pi(gh) = \pi(g) \pi(h)$? In general, no! We only know that, for all $g,h$, there is $\lambda (g,h)$ such that $\pi(gh) = \lambda(g,h) pi(g)\pi(h)$. Computing $\pi(ghk)$ in several ways gives us that $\lambda$ is a cocycle. Assume there is $\mu : G\rightarrow k^*$ such that if we denote, for all $g$, $\pi'(g) := \mu(g) \pi(g)$, we find that $\lambda$ is a coboundary. If $\lambda$ is a cocycle, but not a coboundary, our quest is hopeless: there is no representation giving rise to $\rho$.
The study of whether all cocycles are coboundaries is called cohomology and is quite a long story.
All this being said, I shall try to correct some misunderstandings I think you have, based on what you wrote.
In the way I described above, given a projective representation $\rho$, with the help of any section $s$, we can build a map $\pi$ and a cocycle $\lambda$ such that $\rho = \mathcal{P} \circ \pi$ and such that, for all $g,h \in G^2$, $\pi(gh) = \lambda(g,h)\pi(g)\pi(h)$. The pair $(\pi, \lambda)$ is just data that you can use to reconstruct $\rho$, but is not, strictly speaking, equal to $\rho$. Moreover, as I explained, there are always a lot of pairs $(\pi,\lambda)$ that describe the same projective representation $\rho$! Only in the case where the $\pi$ is a representation and where the $\lambda$ is constant, equal to $1$ can we choose a preferred description of $\rho$.
I should also add that it all the math texts I've read, one should never assume that a map is multi-valued. I think a lot of mathematicians hate multi-valued maps, and such maps should always be (and, as far as I know, are) introduced explicitly as such.
Finally, it seems that in one of the equations that confuse you, you have cancelled out operators on the left and on the right, leaving an operator on the left and only a scalar on the right. This is allowed if the scalar is a short way to write the identity operator times this scalar.
I think the good way of thinking all this is to understand the real definition of a projective representation, then understand that what you thought was the definition is only a description of projective representations; there are also highly non-unique, non-canonical ones.