Lecture 3: Resolutions

Teacher: Benjamin Hennion

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4. Long exact sequences

In this section, we assume that $𝒞$ is an abelian full subcategory of $Mod_A$ (of left $A$-modules) for $A$ an associative ring.

Upshot: $X ∈ 𝒞$, we can tell about elements of $X$.

Theorem (Freyd Mitchell): “Locally”, you can always embed any abelian category in a category of the form $Mod_A$

Snake’s lemma: cf Shapira’s lecture notes

Proof: Let $x ∈ \ker \, h$.

$∃ y ∈ M_2$ st $x=py$

We get

q(g(y)) = h(p(y)) = h(x) = 0

hence $g(y) ∈ N_1$

Set δ(x) = [g(y)] ∈ \coker f

Independent on the choices: take another $y’$ st $p(y’)=p(y)=x$. Replace $y$ by $y-y’$, we can assume $x=0$.

We find $z ∈ M_1$ st $i(z)=y$.

Hence $\underbrace{[g(y)]}_{= [f(z)] = 0} ∈ \coker f$

$\ker δ$: Fix a $x ∈ \ker h$.

δ(x) = 0 ⟺ ∀ y \; \mid \; p(y) = x, [g(y)] = 0 ∈ \coker f\\ ⟺ ∀ y \; \mid \; p(y) = x, ∃ z ∈ M_1 \; \mid \; g(i(z)) = g(y)\\ ⟺ ∀ y \; \mid \; p(y) = x, ∃ z ∈ M_1 \; \mid \; y - i(z) ∈ \ker g\\ ⟺ x ∈ \Im(\ker g ⟶ \ker h)

Same thing for the exactness at $\coker f$.


Theorem: Let

0 ⟶ M_1^\bullet ⟶ M_2^\bullet ⟶ M_3^\bullet ⟶ 0

be an exact sequence in $C(𝒞)$.

Then we have a natural long exact sequence:

⋯ ⟶ H^{n-1}(M_3^\bullet) ⟶ H^{n}(M_1^\bullet) ⟶ H^{n}(M_2^\bullet) ⟶ H^{n}(M_3^\bullet) ⟶ H^{n+1}(M_1^\bullet) ⟶ ⋯

Lemma: $M ∈ C(𝒞)$.

0 ⟶ \underbrace{H^n(M^\bullet) }_{(1)}⟶ \underbrace{\coker d^{n-1}}_{(2)} ⟶ \underbrace{\ker d^{n+1}}_{(3)} ⟶ \underbrace{H^{n+1}(M^\bullet)}_{(4)} ⟶ 0

is exact.

d^{n-1}: M^{n-1} ⟶ M^n\\ d^{n+1}: M^{n+1} ⟶ M^{n+2}\\

Proof: cf pictures

IV. Resolutions

$𝒞$ is an abelian category.

1. Injective and projective objects

An object $I ∈ 𝒞$ is injective:

if the functor \Hom_𝒞(-, I): 𝒞^{op} ⟶ Ab is exact

0 ⟶ M ⟶ N ⟶ P ⟶ 0\\ 0 ⟶ \Hom_𝒞(P, I) ⟶ \Hom_𝒞(N, I) ⟶ \Hom_𝒞(M, I)

Equivalently: $I$ is injective iff $∀ i: M \hookrightarrow N$ monomorphism, $∀f: M ⟶ I, ∃ g: N ⟶ I$ st $gi = f$.

\begin{xy} \xymatrix{ 0 \ar[r] & M \ar[r]^{i} \ar[d]_{ f } & N \ar@{.>}[dl]^{ ∃ \, g } \\ & I & } \end{xy}
An object $P ∈ 𝒞$ is projective:

if the functor \Hom_𝒞(P, -): 𝒞 ⟶ Ab is exact

Equivalently: $P$ is projective iff $∀ M ↠ N$ epimorphism, $∀f: P ⟶ N, ∃ g: P ⟶ M$ st

\begin{xy} \xymatrix{ M \ar[r] \ar@{<.}[d]_{ ∃ } & N \ar@{<-}[dl] \ar[r] & 0\\ P & } \end{xy}

NB: $I ∈ 𝒞$ is injective ⟺ $I ∈ 𝒞^{op}$ is projective


Lemma: Let 0 ⟶ X ⟶ Y ⟶ Z ⟶ 0 be an exact sequence.

If $X$ is injective (resp. $Z$ is projective), the sequence splits (we have a retract (resp. a section)).

In particular, ay additive functor preserves its exactness:

  • if $X$ and $Y$ are injective, so is $Z$
  • if $X$ and $Z$ are injective, so is $Y$

Same for projective objects.

Examples: Let $A$ be a ring

  • Free $A$-modules (generated by a set) are projective
  • Let $P ∈ Mod_A$.

    P \text{ projective } ⟺ ∃ M ∈ Mod_A \; \mid \; P ⊕ M \text{ is a free } A\text{-module}
  • Any projective module is flat.

Proofs:

  • (1): easy
  • (2): $P ⊕ M$ free ⟺ $P ⊕ M$ projective

    cf. picture

    $⟹$: Assume $P$ is projective. Let $E$ be a set of generators of $P$.

  • (3):

    • Any free $A$-module $A^{(E)}$ is flat:

      Fix 0 ⟶ M ⟶ N ⟶ Q ⟶ 0 exact.

      Then

      A^{(E)} ⊗ -: 0 ⟶ M^{(E)} ⟶ N^{(E)} ⟶ Q^{(E)} ⟶ 0 exact

    • P \hookrightarrow M ⊕ P P if $(M ⊕ P) ⊗_A -$ is exact, then so is $P ⊗_A -$


Let $𝒞$ be an abelian category. We say that

$𝒞$ has enough injectives:

if ∀X ∈ 𝒞, ∃ X \hookrightarrow I \text{ a mono, with } I \text{ injective}

$𝒞$ has enough projectives:

if ∀X ∈ 𝒞, ∃ P ↠ X \text{ an epi, with } P \text{ projective}

Theorem: Let $A$ be a ring. The category $Mod_A$ has enough projectives and enough injectives.

Proof: $M ∈ Mod_A$, $E$ set of generators.

A^{(E)} ↠ M ⟶ 0

with $A^{(E)}$ projective.

Examples of injective modules

We don’t care about them

  • An abelian group is injective iff it’s divisible (any element can be divided by an integer):

    $ℚ/ℤ$ is injective in $Mod_ℤ$

  • $A$ commutative: $M ∈ Mod_A$

    M^V ≝ \Hom_{Ab}(M, ℚ/ℤ) ∈ Mod_A
    • $A^{(E)} ↠ M^V$ projective $A^{(E)}$

    • M ⟶ (M^V)^V ⟶ (A^{(E)})^V
      • $(A^{(E)})^V$ is injective (follows from (1))
      • $M ⟶ (A^{(E)})^V$ is a mono

2. Resolutions

Let $x ∈ 𝒞$, where $𝒞$ is abelian.

An injective resolution of $x$:

is a complex 0 ⟶ I^0 \overset{d^0}{⟶} I^1 ⟶ ⋯

and a morphism $φ: X ⟶ I^0$ st

  • $X \overset{φ}{⟶} I^0 \overset{d^0}{⟶} I^1$ is the $0$ morphism
  • every $I^n$ is injective in $𝒞$
  • $X$ identifies with $\ker d^0$ through $φ$ + exact in every $I^n$, for $n > 0$

NB: can reformulate the last assumption

\begin{xy} \xymatrix{ 0 \ar[d] \ar[r] & X \ar[r] \ar[d] & 0 \ar[d] \\ 0 \ar[r] & I^0 \ar[r] & I^1 \ar[r] & ⋯ } \end{xy}

is a quasi-isomorphism.

Dually: a projective resolution of $X ∈ 𝒞$ is a complex $P^\bullet$ and a quasi-isomorphism

\begin{xy} \xymatrix{ ⋯ \ar[r] & P^{-1} \ar[d] \ar[r] & P^0 \ar[d] \ar[r] & 0\\ & 0 \ar[r] & X \ar[r] & 0 \\ } \end{xy}

such that every $P^n$ is projective in $𝒞$

NB: An injective resolution amounts to an exact complex:

0 ⟶ X ⟶ I^0 ⟶ I^1 ⟶ I^2 ⟶ ⋯

with $I^n$ injective $∀n$

Lemma:

(I). Let

0 ⟶ X^0 ⟶ X^1 ⟶ X^2 ⟶ ⋯

be an exact complex, and

0 ⟶ I^0 ⟶ I^1 ⟶ I^2 ⟶ ⋯

a complex of injective objects, not necessarily exact, st

\begin{xy} \xymatrix{ 0 \ar[r] & X^0 \ar[d]_f \ar[r] & X^1 \ar[r] \ar[d]_f & X^2 \ar[d]_f \ar[r] & ⋯ \\ 0 \ar[r] & I^0 \ar[r] & I^1 \ar[r] & I^2 \ar[r] & ⋯ } \end{xy}

Then $f$ is homotopic to $0$.

(II). If $I^\bullet$ is exact, then it is homotopic to $0$.

Proof: cf picture


Theorem: Functoriality of resolution

(i) Assume $𝒞$ has enough injectives (resp. projectives), then any object $x ∈ 𝒞$ admits an injective (resp. projective) resolution.

(ii) Let

  • $f: X ⟶ Y ∈ 𝒞$
  • $0 ⟶ X^0 ⟶ X^1 ⟶ ⋯$ be a resolution (not necessarily injective) of $X$
  • $0 ⟶ I^0 ⟶ I^1 ⟶ ⋯$ be a complex of injectives with a map $Y ⟶ I^0$ st $Y ⟶ I^0 ⟶ I^1$ vanishes.

Then $f$ extends to a morphism $f^\bullet: X^\bullet ⟶ I^\bullet$ that commutes with the maps $X ⟶ X^0$ and $Y ⟶ I^0$

Moreover, any two such maps are homotopic.

Proof: (i). By induction.

Assume we have

0 ⟶ X ⟶ I^0 ⟶ I^1 ⟶ ⋯ ⟶ I^n

exact.

Let $Y = \coker(I^{n-1})$ and let $I^{n+1}$ be an injective st $Y \hookrightarrow I^{n+1}$.

We find 0 ⟶ X ⟶ I^0 ⟶ ⋯ ⟶ I^{n+1} exact.

(ii). By induction:

\begin{xy} \xymatrix{ 0 \ar[r] & X \ar[d]_f \ar[r] & X^0 \ar[d]_f \ar[r] & X^1 \ar[r] \ar[d]_f & ⋯ \ar[r] & X^{n-1} \ar[d]_f \ar[r] & X^{n} \ar[d]_f \ar[r] & ⋯ \\ 0 \ar[r] & Y \ar[r] & I^0 \ar[r] & I^1 \ar[r] & ⋯ \ar[r] & I^{n-1} \ar[r] & I^n \ar[r] & ⋯ } \end{xy}

cf. picture

Corollary: Let $K(\underbrace{\Inj \, 𝒞}_{\rlap{\text{category of injectives}}})$ be the additive category whose

  • objects are complexes 0 ⟶ I^0 ⟶ I^1 ⟶ ⋯ of injective objects

  • and $∀ I^\bullet, J^\bullet$ as above:

    \Hom_{K(\Inj \, 𝒞)}(I^\bullet, J^\bullet) = \Hom_{C(𝒞)}(I^\bullet, J^\bullet)/\sim \; = \; H^0(\Hom_{C(𝒞)}^\bullet(I^\bullet, J^\bullet))

    Taking an injective resolution defines an additive functor:

    Res_I: 𝒞 ⟶ K(\Inj \, 𝒞)

    and an isomorphism of functors

    Id_{𝒞} ≅ H^0 \circ Res

NB: you can do the same for projective resolutions:

K^{≤ 0}(Proj) \ni ⋯ ⟶ P^{-1} ⟶ P^0 ⟶ 0

and we get

Res_P: 𝒞 ⟶ K(Proj(𝒞))

V. Derived functors

1. Definitions

Let $F: 𝒞 ⟶ 𝒟$ be a left exact functor ($𝒞, 𝒟$ abelian). Assume $𝒞$ has enough injectives.

Fix $R_I: 𝒞 ⟶ K(\Inj \; 𝒞)$ a resolution functor.

  • $F$ is additive hence it preserves the homotopy relation: C(𝒞) \ni f \sim g ⟹ F(f) \sim F(g) ∈ C(𝒟)

    F: K(𝒞) ⟶ K(𝒟)

    where

    • objects of $K(𝒞)$ = objects of $C(𝒞)$
    • $\Hom_{K(𝒞)}(X^\bullet, Y^\bullet) = \Hom_{C(𝒞)}(X^\bullet, Y^\bullet)/\sim$
    𝒞 \overset{R_I}{⟶} K(\Inj \, 𝒞) \hookrightarrow K(𝒞) \overset{F}{⟶} K(𝒟) \overset{H^n}{⟶} 𝒟
    The $n$-th right derived functor of $F$:

    is the composite R^n \, F \, ≝ \, H^n \circ F \circ R_I: 𝒞 ⟶ 𝒟

  • Let $G: 𝒞 ⟶ 𝒟$ right exact. Assume $𝒞$ has enough projectives. Fix a projective resolution functor R_P: 𝒞 ⟶ K(\Proj \; 𝒞)

    The $n$-th left derived functor of $G$:

    is the composite L_n \, F \, ≝ \, H^{-n} \circ G \circ R_P: 𝒞 ⟶ 𝒟

In practice: Let $X ∈ 𝒞$.

  • Find an injective resolution of $X$

    0 ⟶ X ⟶ I^0 ⟶ I^1 ⟶ ⋯

    Apply $F$ and take the cohomology:

    R^n \; F = H^n(F(I^\bullet))
  • Find an projective resolution of $X$

    ⋯ ⟶ P^{-1} ⟶ P^0 ⟶ X ⟶ 0

    Apply $G$ and $H^{-n}$:

    L_n \; G = H^{-n}(G(P^\bullet))

Lemma: $X ∈ 𝒞, F: 𝒞 ⟶ 𝒟$.

  • If $F$ is left exact, then

    • $R^n \, F = 0$ for $n < 0$
    • $R^0 \, F ≃ F: 𝒞 ⟶ 𝒟$
  • If $G$ is right exact then

    • $L_n \, G = 0$ for $i < 0$
    • $L_0 \, G ≃ G$

Proof: $F$ left exact ⟺ $F$ preserves kernels

R^0 F(X) = H^0(F(I^0)) = \ker(F(I^0) ⟶ F(I^1)) = F(\ker(I_0 ⟶ I_1)) = F(X)

Properties

  • $R^n \, F$/$L_n \, G$ are additive
  • If $F$ is exact, then $R^n \, F = 0$ for all $n ≠0$
  • If $X$ is injective (resp. projective) then

    • $R^n \, F (X) = 0$ for all $n ≠0$
    • $L_n \, G (X) = 0$ for all $n ≠0$

Theorem: Let

0 ⟶ X ⟶ Y ⟶ Z ⟶ 0 be a short exact sequence in $𝒞$.

We have a long exact sequence

0 ⟶ FX ⟶ FY ⟶ FZ ⟶ R^1 F X ⟶ R^1 F Y ⟶ R^1 F Z ⟶ R^2 FX ⟶ ⋯
⋯ ⟶ L_2 G Z ⟶ L_1 G X ⟶ L_1 G Y ⟶ L_1 G Z ⟶ GX ⟶ GY ⟶ GZ ⟶ 0

Needed Lemma: There exist injective resolutions of $I_X^\bullet, I_Y^\bullet$ and $I_Z^\bullet$ respectively which fit into a commutative diagram:

\begin{xy} \xymatrix{ 0 \ar[r] & X \ar[d] \ar[r] & Y \ar[r] \ar[d] & Z \ar[d] \ar[r] & 0 \\ 0 \ar[r] & I_X^\bullet \ar[r] & I_Y^\bullet \ar[r] & I_Z^\bullet \ar[r] & 0 } \end{xy}

where the rows are exact (in $C(𝒞)$).

Proof: Fix $I_X^\bullet$ and $I_Z^\bullet$ resolutions of $X$ and $Z$.

Let $I_Y^\bullet = I_X^\bullet ⊕ I_Z^\bullet$

To do:

  1. $Y ⟶ I_Y^0$ st $Y ⟶ I_Y^0 ⟶ I_Y^1$ vanishes
  2. $Y = \ker(I_Y^0 ⟶ I_Y^1)$
  3. $H^n(I_Y^\bullet) = 0$ for all $n ≥ 1$
  4. $I_Y^n$ injective

(3) and (4) are obvious.

(1):

\begin{xy} \xymatrix{ 0 \ar[r] & X \ar[d] \ar[r] & Y \ar[r] \ar[d] & Z \ar[d] \ar[r] & 0 \\ 0 \ar[r] & I_X^0 \ar[r] & I_Y^0 = I_X^0 ⊕ I_Z^0 \ar[r] & I_Z^0 \ar[r] & 0 } \end{xy}
  • $Y ⟶ I_Z^0$: take $Y ⟶ Z ⟶ I_Z^0$
  • $∃ Y ⟶ I_X^0$ since $X ⟶ Y$ is a mono and $I_X^0$ is injective

    cf. picture

Proof of the theorem: By the lemma, we have a (split) exact sequence

0 ⟶ I_X^\bullet ⟶ I_Y^\bullet ⟶ I_Z^\bullet ⟶ 0

We find a split exact sequence:

0 ⟶ F(I_X) ⟶ F(I_Y) ⟶ F(I_Z) ⟶ 0

Then, apply a previous theorem to get a long exact sequence:

⟶ H^n F(I_X^\bullet) ⟶ H^n F(I_Y^\bullet) ⟶ H^n F(I_Z^\bullet) ⟶ H^{n+1} F(I_X^\bullet) ⟶ ⋯

So we’re done:

R^n F(X) = H^n (F I_X^\bullet)

2. Examples

$Tor$

Fix $A$ a ring (commutative) and $M ∈ Mod_A$.

The functor $G = - ⊗_A M: Mod_A ⟶ Mod_A$ is right exact.

Tor_n^A(-, M) \; ≝ \; L_n(- ⊗_A M) = L_n(G)

In practice: take $N ∈ Mod_A$, find a projective resolution

⋯ ⟶ P^{-1} ⟶ P^0 ⟶ N ⟶ 0

Compute the cohomology of

⋯ ⟶ P^{-1} ⊗_A M ⟶ P^0 ⊗_A M ⟶ 0

to get $Tor_n^A(N, M)$


Start with

0 ⟶ N_1 ⟶ N_2 ⟶ N_3 ⟶ 0

exact. Get

⋯ ⟶ Tor_2^A(N_3, M) ⟶ Tor_1^A(N_1, M) ⟶ Tor_1^A(N_2, M) ⟶ Tor_1^A(N_3, M) ⟶ N_1 ⊗_A M ⟶ N_2 ⊗_A M ⟶ N_3 ⊗_A M ⟶ 0

Example: $k$ a field. $A = k[x], k ∈ Mod_A$

What is $Tor_n^A(k,k)$?

Resolve $k$ as an $A$-module:

0 ⟶ A = k[x] \overset{μ_x}{⟶} k[x] ⟶ k

is a projective resolution of $k$ as an $A$-module.

Apply $- ⊗_{k[x]} k$ (which amounts to killing $x$):

0 ⟶ \underbrace{k[x] ⊗_{k[x]} k}_{≃ \, k \, ≃ \, Tor_1^A(k,k)} \overset{0}{⟶} \underbrace{k[x] ⊗_{k[x]} k}_{≃ k} ⟶ 0
Tor_0^A(k,k) = k ⊗_{k[x]} k = k

So

Tor_n^A(k,k) = 0 \quad ∀n ≠ 0, 1

Other way to compute it:

0 ⟶ x k[x] ⟶ k[x] ⟶ k ⟶ 0

NB: projective resolutions ⟺ presenting your module with generators ($P^0$) and relations ($P^{-1}$). But then relations are not free themselves ⟶ you iterate.

⋯ ⟶ A^m ⟶ \underbrace{K}_{\text{relations}} ⟶ \underbrace{A^n}_{\text{generators}} ⟶ M

Injective resolutions: the dual notion.

Question: $Tor_n^A(M, N) \text{ vs. } Tor_n^A(N, M)$? These two are the same! (we can do $N ⊗ M$ but $M ⊗ N$ as well)

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