Trying to write the bundle structure for $\Lambda^2(T^*(\mathbb P^3\times \mathbb P^3))$ I'm trying to find an atlas and in general the bundle structure for $\Lambda^2(T^*(\mathbb P^3\times \mathbb P^3))$ respect to, for example, the local chart
$(U_{01},\psi_{01})$, where
$$U_{01}:=\{((x^0:x^1:x^2:x^3),(y^0:y^1:y^2:y^3))\in\mathbb P^3\times \mathbb P^3|x^0\ne 0,y^1\ne 0\}$$
and $\psi_{01}:U_{01}\subset\mathbb P^3\times \mathbb P^3\to\mathbb R^3\times \mathbb R^3$ is the standard chart that divides for the non-zero component.
In particular I was trying to build the bundle structure for $\Lambda^2(T^*(\mathbb P^3\times \mathbb P^3))$ starting from $T^*(\mathbb P^3\times \mathbb P^3)$ and I did this:

Could you tell me if it's correct?
How could I &quot;extend&quot; this idea to build a smooth structure for $\Lambda^2(T^*(\mathbb P^3\times \mathbb P^3))$?
Thank you in advance.
Question: &quot;How could I &quot;extend&quot; this idea to build a smooth structure for Λ2(T∗(P3×P3))? Thank you in advance.&quot;
Answer: An &quot;alternative&quot; approach. Since your bundle is constructed using the pull back of bundles on projective 3-space there is another approach that is more indirect but simpler. It is easier to construct such a structure on $\mathbb{P}^3$ and then pull back this structure to your product $\mathbb{P}^3 \times \mathbb{P}^3$.
Let $X:=\mathbb{P}^3_{k}$ be real ($k:=\mathbb{R}$) projective 3-space and let $Y:=X \times_k X$. There is a formula for the cotagent bundle of a product:
$$\Omega^1_Y \cong p^*\Omega^1_X \oplus q^*\Omega^1_X .$$
There is an exact sequence
$$ 0 \rightarrow \Omega^1_X \rightarrow \mathcal{O}(-1)^{4} \rightarrow \mathcal{O}_X \rightarrow 0.$$
You seek a description of the exterior product
$$\wedge^2 (p^*\Omega_X \oplus q^*\Omega_X)$$
There is (Hartshorne, Ex.II.5.16) a filtration
$$0=F^3 \subseteq F^2 \subseteq F^1 \subseteq F^0:=\wedge^2 (\Omega^1_Y)$$
with
$$F^i/F^{i+1} \cong \wedge^i(p^*\Omega^1_X)\otimes \wedge^{2-i}(q^*\Omega^1_X).$$
This gives an isomorphism
$$C1.\text{  }\wedge^2(T^*_Y):=\wedge^2 (\Omega^1_Y) \cong  $$
$$ p^*(\wedge^2 \Omega^1_X)\oplus q^*(\wedge^2 \Omega^1_X) \oplus p^*\Omega^1_X \otimes q^*\Omega^1_X $$
since pull back commutes with wedge products. Hence if you can construct charts for $\Omega^1_X$, you get charts for $\wedge^2(T^*_Y)$ using formula $C1$ and the projection maps $p,q$.
Exmaple: If $x_0,x_1$ are homogeneous coordinates on $C:=\mathbb{P}^1_k$ you get local coordinates $t:=x_1/x_0$ on $U_0$  and $s:=x_0/x_1:=1/t$ on $U_1$.
The cotangent module $\Omega^1_C$ is on $U_0$ a free module (a trivial bundle) on the basis $dt$. On the open set $U_1$ it has the basis $ds:=d(1/t)=-1/t^2dt$. Hence on the intersection $U_{01}:=U_0 \cap U_1$ you get
$$ds=-\frac{1}{t^2} dt:=\psi_{10}(t,s)dt.$$
The function $\psi_{10}(t,s)$ is invertible on $U_{01}$ and defines the coordinate transformation on this open set. Since $\Omega^1_{C/k}$ is an invertible sheaf, the above calculation proves that
$$ \Omega^1_{C/k} \cong \mathcal{O}(-2).$$
You may work with the real rank one vector bundle on $C$ or the locally trivial sheaf of rank one. The above construction uses the language of &quot;locally trivial sheaves&quot;. There are projection maps
$$p,q: C \times C \rightarrow C$$
and you get the formula
$$\Omega^1_{C\times C/k} \cong p^*\mathcal{O}(-2)\oplus q^*\mathcal{O}(-2)$$
and
$$\wedge^1(\Omega^1_{C\times C/k}) \cong p^*\mathcal{O}(-2)\otimes q^*\mathcal{O}(-2).$$
You get a similar calculation for $X$ and $Y$.
Question: &quot;How could I &quot;extend&quot; this idea to build a smooth structure for Λ2(T∗(P3×P3))? Thank you in advance.&quot;
Answer: An &quot;alternative&quot; approach. Since your bundle is constructed using the pull back of bundles on projective 3-space there is another approach that is more indirect but simpler. It is easier to construct such a structure on $\mathbb{P}^3$ and then pull back this structure to your product $\mathbb{P}^3 \times \mathbb{P}^3$.
Let $X:=\mathbb{P}^3_{k}$ be real ($k:=\mathbb{R}$) projective 3-space and let $Y:=X \times_k X$. There is a formula for the cotagent bundle of a product:
$$\Omega^1_Y \cong p^*\Omega^1_X \oplus q^*\Omega^1_X .$$
There is an exact sequence
$$ 0 \rightarrow \Omega^1_X \rightarrow \mathcal{O}(-1)^{4} \rightarrow \mathcal{O}_X \rightarrow 0.$$
You seek a description of the exterior product
$$\wedge^2 (p^*\Omega_X \oplus q^*\Omega_X)$$
There is (Hartshorne, Ex.II.5.16) a filtration
$$0=F^3 \subseteq F^2 \subseteq F^1 \subseteq F^0:=\wedge^2 (\Omega^1_Y)$$
with
$$F^i/F^{i+1} \cong \wedge^i(p^*\Omega^1_X)\otimes \wedge^{2-i}(q^*\Omega^1_X).$$
This gives an isomorphism
$$C1.\text{  }\wedge^2(T^*_Y):=\wedge^2 (\Omega^1_Y) \cong  $$
$$ p^*(\wedge^2 \Omega^1_X)\oplus q^*(\wedge^2 \Omega^1_X) \oplus p^*\Omega^1_X \otimes q^*\Omega^1_X $$
since pull back commutes with wedge products. Hence if you can construct charts for $\Omega^1_X$, you get charts for $\wedge^2(T^*_Y)$ using formula $C1$ and the projection maps $p,q$.
Exmaple: If $x_0,x_1$ are homogeneous coordinates on $C:=\mathbb{P}^1_k$ you get local coordinates $t:=x_1/x_0$ on $U_0$  and $s:=x_0/x_1:=1/t$ on $U_1$.
The cotangent module $\Omega^1_C$ is on $U_0$ a free module (a trivial bundle) on the basis $dt$. On the open set $U_1$ it has the basis $ds:=d(1/t)=-1/t^2dt$. Hence on the intersection $U_{01}:=U_0 \cap U_1$ you get
$$ds=-\frac{1}{t^2} dt:=\psi_{10}(t,s)dt.$$
The function $\psi_{10}(t,s)$ is invertible on $U_{01}$ and defines the coordinate transformation on this open set. Since $\Omega^1_{C/k}$ is an invertible sheaf, the above calculation proves that
$$ \Omega^1_{C/k} \cong \mathcal{O}(-2).$$
You may work with the real rank one vector bundle on $C$ or the locally trivial sheaf of rank one. The above construction uses the language of &quot;locally trivial sheaves&quot;. There are projection maps
$$p,q: C \times C \rightarrow C$$
and you get the formula
$$\Omega^1_{C\times C/k} \cong p^*\mathcal{O}(-2)\oplus q^*\mathcal{O}(-2)$$
and
$$\wedge^1(\Omega^1_{C\times C/k}) \cong p^*\mathcal{O}(-2)\otimes q^*\mathcal{O}(-2).$$
You get a similar calculation for $X$ and $Y$.