= Solution
Specify a commutative ground <ring> $k$ and a unital associative $k$-<algebra> $R$. For an $R$-<bimodule> $M$, the <Hochschild cochain complex> has
$$
C^q(R,M)=\operatorname{Hom}_k(R^{\otimes_kq},M),\qquad C^0(R,M)=M,
$$
and coboundary
$$
\begin{aligned}
(\delta f)(a_1,\ldots,a_{q+1})={}&a_1f(a_2,\ldots,a_{q+1})\\
&+\sum_{i=1}^{q}(-1)^if(a_1,\ldots,a_ia_{i+1},\ldots,a_{q+1})\\
&+(-1)^{q+1}f(a_1,\ldots,a_q)a_{q+1}.
\end{aligned}
$$
The <bimodule> identities and <associativity> cancel the terms of $\delta^2$ in pairs. The <Hochschild cohomology> is $HH_k^q(R,M)=\ker\delta_q/\operatorname{im}\delta_{q-1}$; the paper's $H^q(R,R)$ uses the regular <bimodule> as coefficients.
Over a <field>, the <bar resolution of an associative algebra> explains its homological meaning. Put $R^e=R\otimes_kR^{\mathrm{op}}$. The degree-$q$ bar term is $R^{\otimes_k(q+2)}$, with the outer factors supplying the $R^e$ action, boundary obtained by multiplying adjacent factors with alternating signs, and augmentation multiplication $R\otimes R\to R$. Inserting $1$ at the left gives a $k$-linear contracting homotopy, proving exactness. The terms are free $R^e$-<modules> after choosing a $k$-basis for the inner tensor factors. Applying $\operatorname{Hom}_{R^e}(-,M)$ gives precisely the displayed cochain complex and therefore
$$
HH_k^q(R,M)\cong\operatorname{Ext}_{R^e}^q(R,M).
$$
This <ground-ring qualification in Hochschild cohomology> matters for an arbitrary <ring>: one can take $k=\mathbb Z$, but the bar terms need not be projective over $R^e$. The ordinary cochain complex then computes relative <Hochschild cohomology>; the absolute Ext identification needs, for example, $R$ projective over $k$. Likewise the extension classification below applies to ground-ring-linearly split extensions. For algebras over a <field> these qualifications are automatic.
In degree zero, $(\delta m)(a)=am-ma$, so \b[$HH^0(R,R)=Z(R)$]. In degree one the cocycle equation is
$$
a\,d(b)-d(ab)+d(a)b=0,
$$
exactly the Leibniz rule for a <derivation of an algebra>. The coboundaries are the <inner derivations> $a\mapsto am-ma$. Thus the <First Hochschild cohomology as outer derivations> is
$$
\boxed{HH^1(R,R)=\operatorname{Der}_k(R)/\operatorname{Inn}(R).}
$$
Its significance is infinitesimal <automorphisms> modulo inner ones. Over the <dual numbers> $k[\varepsilon]/(\varepsilon^2)$, the map $a\mapsto a+\varepsilon d(a)$ is multiplicative precisely when $d$ satisfies the Leibniz rule. Conjugation by $1+\varepsilon m$ gives $a\mapsto a+\varepsilon(ma-am)$, an inner <derivation> with the opposite sign convention. The quotient therefore measures infinitesimal symmetry not arising from conjugation.
Degree two controls extensions and changes of multiplication. A $k$-linear split <square-zero extension of an algebra> with prescribed kernel <bimodule> $M$ can be written as $E=R\oplus M$. Choose a unital section and let $f(a,b)$ be the difference between its product and its value on $ab$. Then
$$
(a,u)(b,v)=(ab,av+ub+f(a,b)).
$$
Computing both parenthesizations of a triple gives the <associativity> condition
$$
a f(b,c)-f(ab,c)+f(a,bc)-f(a,b)c=0,
$$
namely $\delta f=0$. A unital product has $f(1,a)=f(a,1)=0$, so one uses the <normalized Hochschild cochain complex>. Changing section by $g:R\to M$ gives $f\mapsto f+\delta g$. Conversely the displayed product for any normalized two-cocycle constructs such an extension, and a section change gives the corresponding extension isomorphism. This proves that <second Hochschild cohomology classifies split square-zero extensions>:
$$
\boxed{HH^2(R,M)\text{ classifies these extensions, up to equivalence fixing }R,M.}
$$
The zero class is the split algebra extension with no extra product term. With $M=R$, this gives one meaning of the requested $H^2(R,R)$.
For a <first-order associative deformation>, retain the underlying $k$-<module> and put $a*b=ab+\varepsilon f(a,b)$. Expanding $(a*b)*c=a*(b*c)$ modulo $\varepsilon^2$ gives exactly $\delta f=0$. A coordinate change $T=1+\varepsilon g$ transports the product to $T^{-1}(T(a)*T(b))$, whose first-order coefficient is $f+\delta g$. Hence
$$
\boxed{HH^2(R,R)\text{ is the space of inequivalent first-order changes of multiplication.}}
$$
This classification is of infinitesimal deformations, not a guarantee that all classes extend to every order. For $a*b=ab+tf(a,b)+t^2\mu_2(a,b)$ modulo $t^3$, the next <associativity> equation is
$$
\delta\mu_2(a,b,c)=f(f(a,b),c)-f(a,f(b,c)).
$$
Substitution of $\delta f=0$ shows the right-hand side is a three-cocycle. Its class in $HH^3(R,R)$ is the <second-order obstruction to an associative deformation>. It must vanish for a second-order extension to exist. At higher orders the same expansion gives further obstruction cocycles. If $HH^2(R,R)=0$, every first nonzero coefficient of a <formal associative deformation> is a coboundary and can be removed by a coordinate change. Repeating this order by order gives <formal rigidity from vanishing second Hochschild cohomology>, with the transformations converging in the $t$-adic topology. Vanishing of $HH^3$ instead removes these extension obstructions; these are distinct roles.
Examples make the two low degrees concrete. For $R=k[x]$ over a <field>, <inner derivations> vanish and every <derivation> is $h(x)\partial_x$, determined by its value on $x$. The <bimodule> resolution
$$
0\longrightarrow k[x]\otimes k[x]
\xrightarrow{\,x\otimes1-1\otimes x\,}
k[x]\otimes k[x]\longrightarrow k[x]\longrightarrow0
$$
is exact: the middle tensor <ring> is the polynomial <integral domain> $k[u,v]$, multiplication identifies its quotient by $u-v$, and multiplication by $u-v$ is injective. Applying <bimodule> Hom gives zero differential. Consequently the <Hochschild cohomology of a polynomial algebra in one variable> is
$$
HH^0(k[x])=k[x],\qquad HH^1(k[x])=k[x]\partial_x,\qquad HH^q(k[x])=0\quad(q\ge2).
$$
This algebra has many infinitesimal <automorphisms>, while its ordinary associative multiplication is formally rigid.
For $R=k[x,y]$ in characteristic zero, the bilinear map $f(a,b)=(\partial_ya)(\partial_xb)$ is a two-cocycle: substitution in $\delta f$ and the two product rules cancel its four terms. It is not a coboundary, because every $\delta g$ on a commutative <algebra> is symmetric in its two inputs, whereas $f(y,x)-f(x,y)=1$. This class integrates to the <Weyl deformation of a polynomial algebra in two variables>, with $yx-xy=t$. The differential <Ore extension> construction supplies the ordered <basis> $x^iy^j$ over $k[t]$, so the $t$-adic completion has the fixed underlying <module> $k[x,y][[t]]$. Its first multiplication correction is exactly this $f$. This displays a genuinely noncommutative <formal associative deformation> detected by $HH^2$.
Finally, the <Hochschild cup product> $(f\smile g)(a_1,\ldots,a_{p+q})=f(a_1,\ldots,a_p)g(a_{p+1},\ldots,a_{p+q})$ descends to the graded cohomology algebra, and the <Gerstenhaber bracket> organizes the interaction of <derivations> and <formal associative deformation> cocycles. The calculations above explain why the first two positive degrees are especially useful: $HH^1$ measures outer infinitesimal symmetries, while $HH^2$ measures split extensions and infinitesimal associative structures, with degree three detecting the first obstruction to continuing them.
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