# 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?

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Should intuition be added? – user21436 Feb 14 '12 at 0:21
Do you have an intuition about equivalence relations in general? – Qiaochu Yuan Feb 14 '12 at 0:26
@Kannappan: Seems reasonable; done. – Brian M. Scott Feb 14 '12 at 0:27
Thank You @Brian – user21436 Feb 14 '12 at 0:28

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.

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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.

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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 if $(1 + x^2)$). 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.

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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.

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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.

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