Let $G$ be a (connected) reductive group over a perfect field $k$, and let $\xi\in H^1(k,G)$ be a cohomology class.
By a theorem of Steinberg (Serre, Galois cohomology, Appendix 1 to Chapter III, Theorem 11.1),
*if $G$ is quasi-split,* then there exists a $k$-torus $i\colon T\hookrightarrow G$ such that
$$\xi\in i\_\* \big(H^1(k,T)\big).$$



> 
> **Question.** What is an example of $k$, a non-quasi-split reductive $k$-group $G$, and $\xi\in H^1(k,G)$, such that $\xi$ *does not come* from a $k$-torus?
> 
> 
> 


**Remark.** Such $k$ cannot be $\Bbb R$, a $p$-adic field or a number field.
Indeed, let $G$ be a reductive $k$-group, *not necessarily quasi-split*.
If $k$ is $\Bbb R$ or a $p$-adic field, then there exists a $k$-torus $i\colon T\hookrightarrow G$
such that the map
$$i\_\*\colon H^1(k,T)\to H^1(k,G) $$
is surjective.
If $k$ is a number field, then for any finite set $\Xi=\{\xi\_1,\dots,\xi\_n\}\subset H^1(k,G)$ there exists a $k$-torus
$i\colon T\hookrightarrow G$ such that $\Xi\subset i\_\*\big( H^1(k,T)\big)$.
For the last assertion, see M. Borovoi, Abelian Galois cohomology of reductive groups, Mem. Amer. Math. Soc. 132 (1998), no. 626,
Theorem 5.10.



Galois cohomology class of a reductive group not coming from a torus
$\newcommand{\la}{\langle}\newcommand{\ra}{\rangle}$The following example is due to Vladimir Chernousov (private communication).


Let $K={\Bbb Q}(x,y,x',y')$, where $x,y,x',y'$ are variables.
Consider the quadratic forms over $K$
$$ f= \la x,y,-xy\ra\qquad\text{and}\qquad f'=\la x',y',-x'y'\ra.$$
Here $ \la x,y,-xy\ra$ denotes the quadratic form $$f(t\_1,t\_2,t\_3)=xt\_1^2+yt\_2^2-xyt\_3^2.$$



> 
> **Lemma 1.** The quadratic form over $K$
> $$ f-f'=\la x,y,-xy, -x',-y', x'y'\ra$$
> is anisotropic (does not represent 0 nontrivially).
> 
> 
> 


*Proof.* The quadratic form $f-f'$ is monomial and multiplicity free
in the sense of Section 4 of Vladimir Chernousov and Jean-Pierre Serre,
*Lower bounds for essential dimensions via orthogonal representations,*
J. Algebra 305 (2006), no. 2, 1055–1070.
By their Proposition 5 on page 1061, the quadratic form $f-f'$ is anisotropic,
as required.


Consider the quaternion $K$-algebras
$$ D=(x,y)\qquad \text{and} \qquad D'=(x',y').$$
Here $(x,y)$ denotes the 4-dimensional associative $K$-algebra with generators $i,j$ and relations
$$i^2=x, \quad j^2=y,\quad ij=-ji.$$
The reduced norm form for $D$ is isomorphic to $\ \la 1\ra-f=\la 1,-x,-y,xy\ra$.



> 
> **Lemma 2.** The algebraic $K$-groups $G={\rm PGL}(1,D)$ and $G'={\rm PGL}(1,D')$ have no isomorphic maximal tori.
> 
> 
> 


*Proof.*
Assume for the sake of contradiction that $G$ and $G'$ have isomorphic maximal tori.
Then $D$ and $D'$ have isomorphic maximal subfields, say, $L\subset D$ and $L'\subset D'$.
Write $L=K(\sqrt{a})\subset D$ for $a\in K$.
This implies that there exists a pure quaternion $t\_1 i+t\_2 j+t\_3ij$ such that
$$xt\_1^2+yt\_2^2-xyt
\_3^2=a.$$
Thus the quadratic form $f$ represents $a$.
Similarly, from $L'=K(\sqrt{a})\subset D'$ (with the same $a$) we obtain that $f'$ represents $a$.
Therefore, the quadratic form $f-f'$ represents 0 nontrivially, which contradicts Lemma 1.



> 
> **Theorem.** For $G$ and $G'$ as above, let $c\in Z^1(K,G)$ is a 1-cocycle such that $\_c G\simeq G'$.
> Then the cohomology class $[c]\in H^1(K,G)$ does not come from a maximal torus of $G$.
> 
> 
> 


*Proof.*
Assume for the sake of contradiction that there exists a maximal torus
$$ j\colon T\hookrightarrow G\quad\text{such that }\quad [c]\in j\_\* H^1(K,T).$$
Then there exists an embedding
$$j'\colon T\hookrightarrow {}\_c G\simeq G'.$$
Thus $G$ and $G'$ have isomorphic maximal tori, which contradicts Lemma 2.


