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CT groups

I want to describe the equivalence of the category of {$(φ, χ)$}-representations of the {$CT$} group and the category of the graded representations of the corresponding Clifford algebra.

Resources

Gregory Moore. Linear Algebra User’s Manual.

Definitions

A function {$f:V\rightarrow W$} between two complex vectors is antilinear if {$f(x+y)=f(x)+f(y)$} and {$f(sx)=\bar{s}f(x)$} for all vectors {$x,y\in V$} and every complex number {$s\in\mathbb{C}$}.

Any vector space over {$κ = \mathbb{C}$} is, a fortiori also a vector space over {$κ = \mathbb{R}$}. Let us call it {$V_\mathbb{R}$}. It is the same set, but now the vector space structure on this Abelian group is just defined by the action of real scalars. Then we will see that: {$\textrm{dim}_\mathbb{R}V = 2\;\textrm{dim}_\mathbb{C}V$}

Given complex vector space {$V$}, a real structure on {$V$} is an antilinear map {$\mathcal{C}:V\rightarrow V$} such that {$\mathcal{C}^2=+1$}. We further define {$V_+$} to be the vectors such that {$\mathcal{C}(v)=v$} and we call them the "real vectors". We define {$V_-$} to be the vectors such that {$\mathcal{C}(v)=-v$} and we call them the "imaginary vectors". We claim {$V_\mathbb{R}=V_+\oplus V_-$}. {$V_\mathbb{R}$} is the underlying real vector space, the complex vector space {$V$} understood as a real vector space of twice the dimension. This is important so we can represent antilinear operators. Multiplication by {$i$} defines an isomorphism of real vector spaces: {$V_+\cong V_-$}. Thus we have {$\textrm{dim}_\mathbb{R} V_+ = \textrm{dim}_\mathbb{C} V$}.

What can we say about the endomorphisms {$\textrm{End} V_\mathbb{R}$}?

A {$\mathbb{Z}_2$} graded vector space {$V=V_0\oplus V_1$} over {$K$} has graded dimension {$(m|n)$} where {$m$} is the dimension of {$V_0$} and {$n$} is the dimension of {$V_1$}, all over {$K$}. We can write, for example, {$\mathbb{R}^{1|0}$} and {$\mathbb{R}^{0|1}$}.

Note that the graded tensor product switches the sign of 1 out of 4 combinations. This is because noncommutativity is naturally taken care of by {$(e_i\hat{\otimes} 1)\cdot (f_j\hat{\otimes} 1)=e_if_j\hat{\otimes} 1$} and {$(1\hat{\otimes} e_i)\cdot (1\hat{\otimes} f_j)=1\hat{\otimes} e_if_j)$} but we need to track anticommutativity in the case {$(e_i\hat{\otimes} 1)\cdot (1\hat{\otimes} f_j)=-(1\hat{\otimes} f_j)\cdot (e_i\hat{\otimes} 1)$}.

A {$\mathbb{Z}_2$}-graded group is a pair {$(G,\phi)$} where {$G$} is a group and {$\phi :G\rightarrow\mathbb{Z}_2$} is a homomorphism.

• Consequently, a {$\phi$}-twisted extension of {$G$} is an extension of the form {$1\rightarrow U(1)\rightarrow G^{\textrm{tw}}\overset{\pi}{\rightarrow} G\rightarrow 1$} where {$G^\textrm{tw}$} is a group such that {$\tilde{g}$} is linear {$\tilde{g}z=z\tilde{g}$} if {$\phi(g)=1$} and {$\tilde{g}$} is anti-linear {$\tilde{g}z=\bar{z}\tilde{g}$} if {$\phi(g)=-1$}, where {$\tilde{g}\in G^{\textrm{tw}}$} is any lift of {$g\in G$} and {$z\in G^{\textrm{tw}}$} is any phase {$|z|=1$}.
• "the group of operators representing the {$G$}-symmetries of a quantum system form an extension, {$G^\textrm{tw}$}, of {$G$} by {$U(1)$}
• the pullback construction defines {$G^\textrm{tw}:=\{(S,g)|\pi(S)=\rho(g)\}\subset\textrm{Aut}_\mathbb{R}(\mathcal{H})\times G$} where {$\pi :\textrm{Aut}_\mathbb{R}(\mathcal{H})\rightarrow\textrm{Aut}_\textrm{qtm}(\mathbb{P}\mathcal{H})$} and {$\rho :G\rightarrow \textrm{Aut}_\textrm{qtm}(\mathbb{P}\mathcal{H})$}
  • {$\textrm{Aut}_\textrm{qtm}(\mathbb{P}\mathcal{H})$} is the group of quantum automorphisms of a system with Hilbert space {$\mathcal{H}$}.
    • A quantum automorphism is a pair of bijective maps with real linear {$s_1:\mathcal{O}\rightarrow\mathcal{O}$} mapping observables (self-adjoint operators) and {$s_2:\mathcal{S}\rightarrow\mathcal{S}$} mapping states such that probability measures are preserved {$P_{s_1(O),s_2(\rho)}=P_{O,\rho}$}.
    • The group can be identified with the group of (suitably continuous) maps {$s:\mathbb{P}\mathcal{H}\rightarrow\mathbb{P}\mathcal{H}$} such that the overlap function (also known as the transition probability function) is invariant under {$s$}, namely {$\textrm{Tr}P_{s(l_1)}P_{s(l_2)}=\textrm{Tr}P_{l_1}P_{l_2}$}
  • {$\textrm{Aut}_\mathbb{R}(\mathcal{H})$} is the group whose elements are unitary and anti-unitary transformations on the Hilbert space {$\mathcal{H}$}

A right supermodule {$E$} over superalgebra {$A$} is a direct sum decomposition (as an abelian group) {$E=E_0\oplus E_1$} such that multiplication by element of {$A$} satisfies {$E_iA_j\subseteq E_{i+j}$}. This is also called a representation. Consequently, even elements are mapped to diagonal block matrices and odd elements are mapped to off diagonal matrices.

A bigraded group {$G$} has a homomorphism {$G\rightarrow\mathbb{Z}_2\times\mathbb{Z}_2$}, which is to say, a pair of homomorphisms {$(\phi,\chi)$} from {$G$} to {$\mathbb{Z}_2$}.

• A {$(\phi,\chi)$}-representation of {$G$} is a complex {$\mathbb{Z}_2$}-graded vector space {$V=V^0\oplus V^1$} and a homomorphism {$\rho :G\rightarrow\textrm{End}(V_\mathbb{R})$} such that

{$\rho(g)=\left\{\begin{matrix} \mathbb{C}-linear & \phi(g)=+1\\ \mathbb{C}-anti-linear & \phi(g)=-1 \end{matrix}\right.$}

{$and$}

{$\rho(g)=\left\{\begin{matrix} \textrm{even} & \chi(g)=+1\\ \textrm{odd} & \chi(g)=-1 \end{matrix}\right.$}

Note that {$V, V^0, V^1$} are complex vector spaces, {$V_\mathbb{R}$} is a real vector space, and that the elements of {$\textrm{End} V_\mathbb{R}$} are identifiable with real matrices but as such they can express complex linear matrices but also complex antilinear matrices, and they can also be understood as even or odd with respect to the breakdown {$V=V^0\oplus V^1$}.

Note that the distinction between linear and anti-linear is also made by {$\chi(g)$} but externally. {$\chi(g)$} arises from Wigner's theorem which brings to attention both unitary and antiunitary operators. In the context of gapped systems given by a pair of Hilbert spaces {$\mathcal{H}=\mathcal{H}^0\oplus\mathcal{H}^1$}, where {$H>0$} on {$\mathcal{H}^0$} and {$H<0$} on {$\mathcal{H}^1$}, then {$\chi(g)$} is even when it is linear {$\chi(g)=+1$} and odd when it is antilinear {$\chi(g)=-1$}, where {$\rho^{tw}(g)H=\chi(g)H\rho^{tw}(g)$}.

{$M_{2,2}$}

Define the group {$M_{2,2}=\left< \bar{T},\bar{C}\; |\; \bar{T}^2=\bar{C}^2=\bar{T}\bar{C}\bar{T}\bar{C}=1\right>\cong\mathbb{Z}_2\times\mathbb{Z}_2$}

{$M_{2,2}$} has 5 subgroups: {$M_{2,2}, \left<\bar{T}\right>, \left<\bar{C}\right>, \left<\bar{T}\bar{C}\right>, \{\}$}

The {$CT$} groups are the {$\phi$}-twisted extensions of {$M_{2,2}$} and its subgroups. They are bigraded groups.

{$CT$} groups

We want to calculate the {$CT$} groups {$X$} which are group extensions, as below, where {$U$} is a subgroup of {$M_{2,2}= \left<\bar{C},\bar{T}\;|\;\bar{C}^2=\bar{T}^2=(\bar{C}\bar{T})^2=1\right>\cong\mathbb{Z}_2\times\mathbb{Z}_2$}.

{$1\rightarrow U(1)\rightarrow X\rightarrow U \rightarrow 1$}

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When {$U=\{1\}$}, the {$CT$} group is {$U(1)$}

Consider the trivial subgroup {$U=\{1\}$} of {$M_{2,2}$}. What is the group extension?

{$1\rightarrow U(1)\overset{f}{\rightarrow}X\overset{g}{\rightarrow} 1 \rightarrow 1$}

{$f$} is injective, thus {$U(1)\cong\textrm{Img}(f)$} but also {$\textrm{Img}(f)=\textrm{Ker}(g)=X$}, thus {$X\cong U(1)$} is the {$CT$} group.

The irreducible complex representations of {$U(1)$} are the rotations {$\phi_n(\theta)=(e^{i\theta})^n$} where {$n\in\mathbb{Z}$} and they are all inequivalent and notably {$\phi_n$} is conjugate to {$\phi_{-n}$}.

The irreducible real representations of {$U(1)$} are the rotations:

{$\rho_n(e^{i\theta})=\begin{pmatrix} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \\ \end{pmatrix} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{pmatrix}^n$}

where {$n\geq 0, n\in\mathbb{Z}$}. When {$n=0$}, this gives the trivial representation. Note that {$\rho_n\cong\rho_{-n}$} by way of a real matrix whereas there is no such complex number that by which {$\phi_n$} would be isomorphic to {$\phi_{-n}$} as they are conjugate.

The complex vector space is {$\mathbb{Z}_2$}-graded.

The trivial complex Clifford algebra {$\mathbb{Cl}_0$} has no generators thus is purely even. As a superalgebra it has two inequivalent irreducible graded modules {$M_0^+\cong\mathbb{C}^{1|0}$} and {$M_0^-\cong\mathbb{C}^{0|1}$}.

Concretely: {$c_1\rightarrow\begin{pmatrix} c_1 & 0 \\ 0 & 0 \\ \end{pmatrix}$} acting on {$\begin{pmatrix} \mathbb{C} \\ 0 \\ \end{pmatrix}$} and {$c_1\rightarrow\begin{pmatrix} 0 & 0 \\ 0 & c_1 \end{pmatrix}$} acting on {$\begin{pmatrix} 0 \\ \mathbb{C} \end{pmatrix}$}

---

{$U=\{1,\bar{C}\bar{T}\}$}

Now consider {$U = \{1, \bar{S}\}$} where {$\bar{S}=\bar{C}\bar{T}$} lifts to {$S$} which is complex linear and odd. The embedding {$\iota (U(1)$} of {$U(1)$} in {$G$} is represented by complex linear matrices. Thus, apparently, {$S$} commutes with {$\iota (U(1)$} and thus with all of {$G$}. The fact that {$\iota (U(1)$} is in the center of {$G$} means that the short exact sequence is a central extension of {$G$}. There is a unique central extension {$G\cong U(1)\times\mathbb{Z}_2$} and it is abelian. We know that {$S^2=\theta\in\iota(U(1))$}, thus we can consider {$b=S\theta^{-\frac{1}{2}}$} and note that {$b^2=1$}. Thus we can suppose {$S^2=1$}. Moreover, {$S$} is {$\mathbb{C}$}-linear, as are all elements of {$G$}, so we can think of them as acting on {$W$} as a vector space over {$\mathbb{C}$}. Note that {$S$} acts on {$W=W^0\oplus W^1$} as an odd operator whereas the elements of {$\iota (U(1))$} are even operators. Therefore, we can take {$V = W$} and identify {$S$} with an odd generator of {$\mathbb{C}\ell_1$}.

{$\mathbb{C}\ell_1$} has a single generator {$e_1$}, which is odd. Thus it has one even dimension and one odd dimension. We show that it has a unique irreducible graded representation {$\mathbb{C}^{1|1}$}. We must represent {$\rho (e_1)$} by an odd operator which squares to {$+1$}. The most general such operator is

{$\rho(e_1)=x\sigma^1 + y\sigma^2=\begin{pmatrix} 0 & x-yi \\ x+yi & 0 \\ \end{pmatrix}$} where {$x^2+y^2=1$}

{$\rho(e_1)=\begin{pmatrix} 0 & e^{-i\alpha} \\ e^{i\alpha} & 0 \\ \end{pmatrix}$}

These operators are equivalent for all {$\alpha$} as we can conjugate by the even transformation {$\cos\theta + i\sin\theta\sigma^3$}.

{$\begin{pmatrix} \cos\theta + i\sin\theta & 0 \\ 0 & \cos\theta - i\sin\theta \\ \end{pmatrix} \begin{pmatrix} 0 & x-yi \\ x+yi & 0 \\ \end{pmatrix} \begin{pmatrix} \cos\theta - i\sin\theta & 0 \\ 0 & \cos\theta + i\sin\theta \\ \end{pmatrix} = $}

{$\begin{pmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \\ \end{pmatrix} \begin{pmatrix} 0 & e^{-i\alpha} \\ e^{i\alpha} & 0 \\ \end{pmatrix} \begin{pmatrix} e^{-i\theta} & 0 \\ 0 & e^{i\theta} \\ \end{pmatrix} = \begin{pmatrix} 0 & e^{i(2\theta - \alpha)} \\ e^{i(\alpha - 2\theta)} & 0 \\ \end{pmatrix}$}

Two possibilities are {$\alpha=0$} and {$\alpha=\frac{\pi}{2}$}, yielding, respectively,

{$\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$} and {$\begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix}$}

Going backwards, given the irreducible graded representation of {$\mathbb{C}\ell_1$}, we take {$W=V$} and {$e_1$} represents {$S$}.

---

{$U=\{1,\bar{C}\}$}

{$C$} is {$\mathbb{C}$} antilinear. Apparently, this means that for {$z\in\iota (U(1))$}, {$Cz=z^{-1}C$}. Thus {$G$} is given by the noncommutative groups {$\textrm{Pin}_+(2)$} when {$T^2=+1$} and {$\textrm{Pin}_-(2)$} when {$T^2=-1$}.

---

When {$U=\{1,\bar{C}\}\cong\mathbb{Z}_2$}, the {$CT$} group is {$\textrm{Pin}_+(2)$} if {$\tilde{C}^2=+1$} and {$\textrm{Pin}_-(2)$} if {$\tilde{C}^2=-1$}

---

When {$U\cong\mathbb{Z}_2$}, the {$CT$} group is {$\textrm{Pin}_+(2)$} or {$\textrm{Pin}_-(2)$}

{$(\phi,\chi)$}-representations of {$CT$}-groups

Identity: linear, even: {$\rho(I)=\begin{pmatrix}1 & 0 \\ 0 & 1 \\ \end{pmatrix}$}

Time reversal, antilinear, even: {$\rho(\bar{T})=\begin{pmatrix}1 & 0 \\ 0 & -1 \\ \end{pmatrix}$}

Charge conjugation, antilinear, odd: {$\rho(\bar{C})=\begin{pmatrix}0 & 1 \\ 1 & 0 \\ \end{pmatrix}$}

Parity, linear, odd: {$\rho(\bar{CT})=\begin{pmatrix}0 & -1 \\ 1 & 0 \\ \end{pmatrix}$}

Perspectives

Time reversal fixes absolute perspective and flips relative perspective.

{$\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}=\begin{pmatrix}x \\ -y \end{pmatrix}$}

Antilinear swaps absolute and relative perspectives.

• First perspective: antilinear, odd, squares to +1: {$\rho(\bar{C})=\begin{pmatrix}0 & 1 \\ 1 & 0 \\ \end{pmatrix}$}
• Second perspective: antilinear, odd, squares to -1: {$\rho(i\bar{C})=\begin{pmatrix}0 & i \\ i & 0 \\ \end{pmatrix}$}
• Third perspective: linear, odd, squares to -1: {$\rho(\bar{C}\bar{T})=\begin{pmatrix}0 & 1 \\ -1 & 0 \\ \end{pmatrix}$}
• Fourth perspective: linear, odd, squares to +1: {$\rho(i\bar{C}\bar{T})=\begin{pmatrix}0 & i \\ -i & 0 \\ \end{pmatrix}$}
• Fifth perspective?: antilinear, even, squares to +1: {$\rho(\bar{T})=\begin{pmatrix}1 & 0 \\ 0 & -1 \\ \end{pmatrix}$}
• Sixth perspective?: antilinear, even, squares to -1: {$\rho(i\bar{T})=\begin{pmatrix}i & 0 \\ 0 & -i \\ \end{pmatrix}$}
• Seventh perspective?: linear, even, squares to -1: {$\rho(iI)=\begin{pmatrix}i & 0 \\ 0 & i \\ \end{pmatrix}$}
• Eighth perspective?: linear, even, squares to +1 {$\rho(I)=\begin{pmatrix}1 & 0 \\ 0 & 1 \\ \end{pmatrix}$}

{$\rho(I)=\begin{pmatrix}1 & 0 \\ 0 & 1 \\ \end{pmatrix}$}

Note that {$i$} is a choice from {$\pm i$} and similarly {$\rho(\bar{T})$} is a choice from {$\pm\rho(\bar{T})$}.

The main (and only) perspective {$\bar{C}$} is the swap of absolute and relative spaces. Then it is a matter of the context for the perspective, which grows more sophisticated.

Thoughts

CT groups are relating structure (two on-and-off switches) and recurring activity (the circle). Bott periodicity cycles through all the ways of relating structure and recurring activity, thus all the contexts for patterns.

Group extensions express a short exact sequence, a division of everything, notably the foursome, how perspectives carve up space and fit together.

Thus we are matching graded representations of Clifford algebras and bigraded representations of CT groups. We also have nongraded Clifford representations. Are these the three minds?

---

@**David Michael Roberts** @**Todd Trimble** I am trying to understand this theorem in Section 2.6 of Gregory Moore's [Quantum Symmetries and K-Theory](https://www.physics.rutgers.edu/~gmoore/PiTP-LecturesA.pdf).

[CTGroups-Figure11.png](/user_uploads/21317/HMmOricv1tqKjhw1kFEdkp8D/CTGroups-Figure11.png)

• Theorem** There is a one-one correspondence, given in the table above, between the ten CT groups and the ten real super-division algebras (equivalently, the 10 Morita classes of the real and complex Clifford algebras) such that there is an equivalence of categories between the (φ, χ)-representations of the CT group and the graded representations of the corresponding Clifford algebra.

Basically, we're associating {$\mathbb{Z}_2\times\mathbb{Z}_2$$-graded group representations with {$\mathbb{Z}_2$$-graded Clifford algebra representations. I want to understand the details but I am still struggling to understand the definitions. Anyways, let me describe the simplest case.

Consider the group extension {$1\rightarrow U(1)\rightarrow G\rightarrow U\rightarrow 1$} when {$U=1$$, the trivial group. Then {$G\cong U(1)$$, the circle group, is the {$CT$} group. Moore writes:

First, consider the subgroup {$U = \{1\}$$. A {$(φ, χ)$} representation {$W$} is simply a {$\mathbb{Z}_2$} -graded complex vector space, so {$V = W$} is a graded {$Cℓ_0$} -module.

What apparently he is saying is that the representation space is given ($$\mathbb{Z}_2$} -graded complex vector space {$V=V^0\oplus V^1$$) and the representation {$\rho :G\rightarrow\textrm{End}(V_\mathbb{R})$} is obvious. But I can't figure out what it is!

According to [Circle group: Representations](https://en.wikipedia.org/wiki/Circle_group), the irreducible real representations of the circle group are the trivial representation and also {$\phi_n, n\in\mathbb{Z}, n> 0$} given by

{$\rho_n(e^{i\theta})=\begin{pmatrix} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \\ \end{pmatrix} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{pmatrix}^n$}

These matrices are all complex linear. Now, if they had to be either even or odd, then by the grading they would all be prohibited, for they are neither even nor odd. We would be left with the {$1$$-dimensional trivial representation {$\rho(g)=(1)$} for all {$g\in G$$. So I suppose that is what must be meant. This means that {$V_\mathbb{R}$} should be understood as {$V_\mathbb{R}=V_+\oplus V_-$} where {$V_+=V^0_\mathbb{R}$} and {$V_-=V^1_\mathbb{R}$$.

Given that we have the {$\mathbb{Z}_2$$-graded complex vector space {$V_\mathbb{R}=V_+\oplus V_-$$, we have two irreducible {$(\phi,\chi)$$-representations of {$G$$, namely:

{$\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix}$} and {$\begin{pmatrix} 0 & 0 \\ 0 & 1 \\ \end{pmatrix}$}

This accords with the graded representations of the complex Clifford algebra {$\mathbb{C}\ell_0$} he describes in [Quantum Symmetries and Compatible Hamiltonians](https://www.physics.rutgers.edu/~gmoore/QuantumSymmetryBook.pdf) in Section 13.2.1 Structure of the (graded and ungraded) algebras and modules. He writes:

Of course {$\mathbb{C}\ell_0\cong\mathbb{C}$} is purely even. Nevertheless, as a superalgebra it has two inequivalent irreducible graded modules {$M_0^+ \cong C^{1|0}$} and {$M_0^−\cong C^{0|1}$$.

Super-modules are defined in QS&CH Section 12.4 Modules over superalgebras.

A super-module M over a super-algebra A (where A is itself a superalgebra over a field κ) is a supervector space M over κ together with a κ-linear map A × M → M defining a left-action or a right-action.

So the map is multiplication of complex numbers as in {$\mathbb{C}\times C^{1|0}\rightarrow C^{1|0}$}

So I suspect I am on track. Going back, in the opposite direction, he says in QS&CH, Section 16. Realizing the 10 classes using the CT groups

For {$\mathbb{C}\ell_0$} we take {$W = V$}

So the two situations are very similar. But for the group we have a trivial representation, whereas for the Clifford algebra, we have multiplication by {$c\in\mathbb{C}$$. At least, that's what I think.

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My intuition, which I doubted for a while, but now return to, is that this is all about breaking a {$2\times 2$} matrix into its 4 components.

{$\begin{pmatrix} 1 & 0 \\ 0 & 1\\ \end{pmatrix}$} is even complex linear, as with identity

{$\begin{pmatrix} 1 & 0 \\ 0 & -1\\ \end{pmatrix}$} is even complex antilinear, as with time reversal {$\bar{T}$}

{$\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$} is odd complex antilinear, as with charge conjugation {$\bar{C}$}

{$\begin{pmatrix} 0 & -1 \\ 1 & 0\\ \end{pmatrix}$} is odd complex linear as with the product, parity {$\bar{S}=\bar{C}\bar{T}$}

If this is on track, then it starts to provide the metaphysical intuition that I am seeking. Time reversal fixes position and flips momentum, which is to say, it distinguishes an absolute frame (for position) and a relative frame (for momentum). Then charge conjugation swaps absolute frame and relative frame, perhaps thereby defining a perspective. My goal is to get metaphysical intuition on these perspectives and then relate that back to the construction described by [Stone, Chiu, Roy](https://arxiv.org/abs/1005.3213) in terms of Lie group embeddings and {$0,1,2,\cdots 7$} mutually anticommuting linear complex structures. I think the later may encode the "divisions of everything" that I have documented, the ways of carving up mental space into perspectives, much like a chain complex or exact sequence carve up conceptual space with homomorphisms.