# What's an intuitive way of looking at quotient spaces?

I understand the concept of $\mathbb{Z}/n\mathbb{Z}$, but I am having a really hard time understanding how this concept of quotients applies to vector spaces. Suppose $V = \mathbb{F}[x]$ is a vector space and $U \le V$. What exactly does $V/U$ represent?

• Do you have an intuition about equivalence relations in general? Feb 14 '12 at 0:26
• – lhf
Sep 22 '16 at 20:17

The way I think about quotient spaces (or quotient algebraic structures in general) is as an identification of things which differ by some subspace/subgroup/subring/... of the original structure. Then the quotient space (resp. algebraic structure) can be thought of as what you get when you squash the subspace (resp. substructure) to a point and extend to the rest of the space (resp. structure).

To illustrate this, with $\mathbb{Z}/n\mathbb{Z}$, we identify integers which differ by a multiple of $n$. What we do is squash $n\mathbb{Z}$ to a point, namely $0$, and this extends to the rest of the space by squashing $a+n\mathbb{Z}$ to $a$ for $0 \le a < n$. Then the operations of addition and so on are inherited naturally from the quotient operation.

As another example, with $\mathbb{R}^2/\langle (1,1) \rangle$ we identify things which lie on the same line lying at $45^{\circ}$ to the (positive) horizontal, i.e. we identify vectors which differ by some scalar multiple of the vector $(1,1)$. We can visualise this as contracting $\langle (1,1) \rangle$ to a point, which when you extend linearly contracts $\mathbb{R}^2$ to a line through the origin, leaving you with $\langle (1,-1) \rangle$. (Imagine squashing the whole plane down towards the origin perpendicular to the line $\langle (1,1) \rangle$). Another perspective is that the 'points' in $\mathbb{R}^2/\langle (1,1) \rangle$ can be thought of as precisely the lines in $\mathbb{R}^2$ with gradient $1$.

More generally, if $V$ is a vector space and $U \le V$ is a subspace then you can think of $V/U$ as the space you get when you identify two elements $v, v' \in V$ if $v'=v+u$ for some $u \in U$. What it 'looks like' is what you get when you contract $U$ to a point and extend linearly to the rest of the space.

I hope this wasn't too waffly.

• Hi! is it possible to explain more detailed your sentence: "what you get when you squash the subspace (resp. substructure) to a point and extend to the rest of the space (resp. structure)." So, if I have $V/W$, I have to squash $W$ to a point - this is ok, but is not clear for me what does it mean "extend to the rest of the space". Thanks!
– Iuli
Nov 12 '19 at 14:23
• @luli: This is a blast from the past (7.5 years ago) so I'm not 100% sure what I meant, but I think this is the gist: if you squish $\langle (1,1) \rangle$ to a point but leave the rest of the plane intact, then what you have just looks like a plane with a line removed. But instead what you have to do is squash every line parallel to $\langle (1,1) \rangle$ (not just the one through the origin) to a point. What results is a line parallel to $(1,-1)$. Nov 12 '19 at 15:33
• thanks for your time. sincerely, the topic is still unclear. from my point of view, the plane $\mathbb{R}^2$ can be covered with parallel lines with $(1,1)$. Why the result is a parallel line with $(1,-1)$?
– Iuli
Nov 13 '19 at 6:50
• @luli: It really is just intuition. The idea is you're collapsing each of the lines that are parallel to $(1,1)$ to a point; I visualise them collapsing onto a line that is perpendicular to $(1,1)$. From a formal perspective this isn't what's happening at all: the 'vectors' in the new vector space are the lines themselves. Nov 13 '19 at 12:54

A vector space quotient is a very simple projection when viewed in an appropriate basis. Namely, any basis of the subspace U may be extended to a basis of the whole space V. Then modding out by U amounts to zeroing out the components of the basis corresponding to U, i.e. projecting onto the complementary subspace formed by all the other components.

For example, consider the vector space $\mathbb R^{\infty}$ of formal power series with real coefficients. Modding out by the subspace of odd series $\mathbb R[x,x^3,x^5,\ldots]$ corresponds to projecting onto the subspace of even power series $\mathbb R[1,x^2,x^4,\ldots]$ i.e. taking the even part of a power series (and vice versa). Combining the two yields the decomposition of a series as the sum of its even and odd parts.

Similarly, modding out by $\mathbb R[x,x^2,x^3,\cdots]$ corresponds to projecting onto the subspace of constant series $\mathbb R[1]$, i.e. "evaluating" at $x = 0$. Modding out by $\mathbb R[x^2,x^3,x^4,\ldots]$ corresponds to projecting onto the "tangent" space $\mathbb R[1,x]\:$ via $\:f(x)\mapsto f(0) + f'(0)\:x =$ first-order Taylor approximant.

I visualize the quotient space as a "foliation" of the space - think salad, but with rectilinear layers extending infinitely outwards - and in this case the layers of the foliation in $V/W$ are cosets of the subspace $W$. Geometrically, the cosets are just the space $W$ shifted in any of the directions normal to it. For example, a line that goes through the origin in $\mathbb{R}^2$ is a subspace, to get the quotient we designate every line parallel to it in the plane as an element of the quotient, and we parametrize these layers by drawing a second, nonparallel line through the first at the origin, so that each point on the second line represents the layer it intersects in the plane. In order to add or scalar-multiply two layers, we simply do the associated vector operations on the subspace given by our second line!

You can do a similar geometry exercise in $\mathbb{R}^3$ with lines and planes. With higher dimensions or more abstract spaces, we don't always have the luxury of a directly visualizable illustration, but I find the idea that we're cutting a space up into salad layers helpful all the same.

• Sorry for necropost: What confuses me is that the cosets, as you said are represented by pairs with fixed y-value, i.e., pairs of the form (x, $y_0$) with $y_0$ fixed. But there are uncountably-many such pairs. But the quotient space here is 1-dimensional. How can this be, if each coset is an element of the quotient and pairs (x,$y_o$), (x, $y_0'$); $y_0 \neq y_0'$ are inequivalent. Also, the quotient $\mathbb R^3/ \mathbb R )$ is 2-dimensional ; it identifies/collapses all points (x,y,z) with last two entries fixed. But each representative is a line, .so 1-dimensional. What am I missing?
– MSIS
Sep 9 '20 at 20:02

Let's look at a simple example. I'll take $V = \mathbb{R^2} = \{(x,y): x,y \in \mathbb{R}\}$ and $U = \{(x,0): x \in \mathbb{R}\}$; i.e $U$ is the $x$-axis inside the plane. What is $V/U$? On an algebraic level, it's the vector space of cosets of $U$: i.e. I identify two points $(x,y)$ and $(x', y')$ if $(x,y) - (x',y') \in U$. In our context, this happens exactly when $y = y'$. So we are identifying all points that have a given $y$-coordinate, and addition of cosets $(x,y) + U$ is just adding the $y$-coordinates. While it might be tempting to identify $U$ with the $y$-axis, this is somewhat misleading, since this would lead you to think that the quotient space $V/U$ is a subspace of $V$, which it is not.

• Sorry for the necropost: What confuses me is that the cosets, as you said are represented by pairs with fixed y-value, i.e., pairs of the form (x, $y_0$) with $y_0$ fixed. But there are uncountably-many such pairs. But the quotient space here is 1-dimensional. How can this be, if each coset is an element of the quotient and pairs (x,$y_o$), (x, $y_0'$); $y_0 \neq y_0'$ are inequivalent.
– MSIS
Sep 9 '20 at 19:46

What $V/U$ represents depends heavily on what $V$ and $U$ represent.

Anyways, since you mention $F[x]$, one classical example is $F = \mathbb{R}$ and $V$ = $\mathbb{R}[x]$, and $U = (1+x^2) V$. (i.e. all multiples of $(1 + x^2)$). Then $V/U \cong \mathbb{C}$, a two-dimensional real vector space. This usually comes up in the context of rings moreso than pure linear algebra, though.

In this case, it's clear what $U$ and $V$ represent; $V$ is meant to represent polynomials in a hypothetical square root of $-1$, and $U$ is meant to represent those polynomials that are intended to be zero.

I like Clive Newstead's answer. That's kinda how I always thought about it. You squash a subspace, call it L (it can be topological and algebraic for sure, but probably more types!), into a point, and this is the "zero point" (when you use a topological structure continuous map, then the set of zeroes of this map are the "zero points" if you are using an algebraic map, it is also an isomorphic map are and this means that in the domain, linear subspaces of a general vector space V get squashed down the literal zero vector in V because they are the kernel of this linear map by assumption, so they are in the same equivalence class of 0 which is of dimension V, and f(0)=0).

You then shift this linear subspace around the space to see what are the equivalence classes of each space, which is explained perfectly in Clive Newstead's paragraph 3. Now, there one important key to me that I might add to Clive's response is really just another way of saying that points in the quotient space are heavy as Clive said in paragraph 2. It is that in the V that I mentioned earlier, points are literally things with no internal structure. They are empty inside. But when taking the quotient space, points become equivalence classes (look at cosets). So every point in V is no longer alone and empty inside. Now, inside, it has all the points that are equivalent to it. So the internal structure of points of V is somehow increased by collapsing V via some nontrivial subset.

Points gaining information seems very much like massive pointlike particles. Also, each point of V that is "equivalent" to some other point in V reflects some invariance of the element, which is what particles experience as they pass through time. Particles look the same over time for all we know. Electrons' shape doesn't change anywhere. Every electron is identical to every other electron (how about all elementary particles?). So choosing different "coset representatives" (eg: different points in V equivalent to each other in terms of how they differ the collapsed subspace L) is sort of like choosing the same particle at different moments in its timelike trajectory existence of General Relativity.

I am motivated by this way of conceptualizing quotient spaces because it seems to correlate with the foundations of physics in some sense...Point like particles get mass at the same instance that a timelike trajectory (which is a curvy line that is intuitively homeomorphic to a plain old coordinate dimension, say t, in spacetime)gets identified into a point.

We don't know how particles get mass in a solid geometry picture, and we don't know how time becomes moments that we see in this world. A small enlarging and a big collapsing. A quotient space. The act of "dividing by zero", or a timelike curve in 4d spacetime, creates motion of massive particles in a 3D space. Also, I suspect that the orientation of integrals that define periods related to homology classes in the math world might say something about this "holy grail" of the quotient of spacetime by time, which in turn might say something about both inertia of massive bodies, and the direction of time.

Just simple quotient spaces. I don't see a clear picture of my ideas, but I know that quotient spaces bring some order to them. It gives me a legitimate way of perceiving quotient spaces and builds intuition for questions that might be meaningful one day.