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$\newcommand{\la}{\langle}\newcommand{\ra}{\rangle}$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 $G\simeq {}_c G$. Then the cohomology class $[c]\in H^1(K,G)$ does not come from a maximal torus in $G$.

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 $$j'\colon T\hookrightarrow {}_c G=G'.$$ Thus $G$ and $G'$ have isomorphic maximal tori, which contradicts Lemma 2.

$\newcommand{\la}{\langle}\newcommand{\ra}{\rangle}$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.

$\newcommand{\la}{\langle}\newcommand{\ra}{\rangle}$The following example is due to Vladimir Chernousov (private communication).

Let $K={\Bbb Q}(x,y,x',y')$, where $x,y,x',v'$ 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 $G\simeq {}_c G$. Then the cohomology class $[c]\in H^1(K,G)$ does not come from a maximal torus in $G$.

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 $$j'\colon T\hookrightarrow {}_c G=G'.$$ Thus $G$ and $G'$ have isomorphic maximal tori, which contradicts Lemma 2.

$\newcommand{\la}{\langle}\newcommand{\ra}{\rangle}$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 $G\simeq {}_c G$. Then the cohomology class $[c]\in H^1(K,G)$ does not come from a maximal torus in $G$.

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 $$j'\colon T\hookrightarrow {}_c G=G'.$$ Thus $G$ and $G'$ have isomorphic maximal tori, which contradicts Lemma 2.

The following example is due to Vladimir Chernousov (private communication).

Let $K={\Bbb Q}(x,y,x',y')$, where $x,y,x',v'$ are variables. Consider the quadratic forms over $K$ $$ f= <x,y,-xy>\qquad\text{and}\qquad f'=<x',y',-x'y'>.$$ Here $ <x,y,-xy>$ 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 $K$ $$ f-f'=<x,y,-xy, -x',-y', x'y'>$$ 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 $\ <1>-f=<1,-x,-y,xy>$.

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 $G\simeq {}_c G$. Then the cohomology class $[c]\in H^1(K,G)$ does not come from a maximal torus in $G$.

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 $$j'\colon T\hookrightarrow {}_c G=G'.$$ Thus $G$ and $G'$ have isomorphic maximal tori, which contradicts Lemma 2.

$\newcommand{\la}{\langle}\newcommand{\ra}{\rangle}$The following example is due to Vladimir Chernousov (private communication).

Let $K={\Bbb Q}(x,y,x',y')$, where $x,y,x',v'$ 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 $G\simeq {}_c G$. Then the cohomology class $[c]\in H^1(K,G)$ does not come from a maximal torus in $G$.

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 $$j'\colon T\hookrightarrow {}_c G=G'.$$ Thus $G$ and $G'$ have isomorphic maximal tori, which contradicts Lemma 2.