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Let $V=V_0\oplus V_1$ be a superspace, where its even subspace $V_0$ and odd subspace $V_1$ both are $k$-dimensional. Let $(e_1,\cdots,e_k|f_1,\cdots,f_k)$ be a basis of $V$.

Consider linear transformations $\varphi,\psi$ on $V$, such that $\varphi(V_0)\subseteq V_0, \varphi(V_1)\subseteq V_1$; $\psi(V_0)\subseteq V_1,\psi(V_1)\subseteq V_0$.

Suppose $\varphi$ is nilpotent and $\varphi\psi-\psi\varphi=\theta$, where $\theta(e_i)=f_i$ and $\theta(f_i)=-e_i$. Then is it true that ${rank}(\varphi_0)={rank}(\varphi_1)$? Here $\varphi_0$ denotes restricting on even subspace , i.e. $\varphi|_{V_0}$.

If $\varphi^2=0$, then we know $\varphi$ anti-commute with $\theta$. Then $\varphi_0\theta_1=-\theta_0\varphi_1$, clearly ${rank}(\varphi_0)={rank}(\varphi_1)$. Is this true in general, when $\varphi^n=0$ with arbitrary $n$?

Thanks!

qinr
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1 Answers1

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$\DeclareMathOperator{\rank}{rank} \require{cancel}$ Let $a \in \ker(\varphi_0)$. Then

$$[\varphi, \psi](a, 0) = \varphi\circ\psi(a, 0) - \cancelto{(0, 0)}{\psi\circ\varphi(a, 0)} = \theta(a, 0)$$

Hence

$$\varphi\circ\psi(a, 0) = \theta(a, 0) = (0, a)$$

Since $\varphi$ is $n$-nilpotent $$\varphi^{n-1}(0, a) = \varphi^{n-1}(\varphi\circ\psi(a, 0)) = \varphi^{n}\circ\psi(a, 0) = (0, 0)$$

This means that $a \in \ker(\varphi^{n-1}_1)$ which implies

$$\ker(\varphi_0) \subseteq \ker(\varphi^{n-1}_1)$$

By the analogous reasoning we are also able to conclude that

$$\ker(\varphi_1) \subseteq \ker(\varphi^{n-1}_0)$$

As you noted, when $n=2$ it is immediate to see that $\ker(\varphi_0) = \ker(\varphi_1)$ whence follow the thesis. However this may not be the case for $n = 3$. Consider for example $V_0 = V_1 = \mathbb{R}^3_{\mathbb{R}}$ with its canonical basis and the linear transformation represented by the upper triangular matrix

$$T = \begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix} $$

Next choose $\varphi = T \oplus T^2$ and any $\psi$ such that $[\varphi, \psi] = \theta$, for instance

$$\psi : V_0 \oplus V_1 \to V_0 \oplus V_1 : (x, y) \mapsto (Ay, Bx)$$

where

$$A = \begin{bmatrix} -24 & -18 & 22 \\ -19 & 6 & -6 \\ 0 & 0 & -48 \\ \end{bmatrix} \qquad% B = \begin{bmatrix} -42 & -12 & -28 \\ 0 & -5 & -23 \\ 0 & -19 & -13 \\ \end{bmatrix}$$

Thus all hypotheses are met but

$$\dim\ker(\varphi_0) = 1 < 2 = \dim\ker(\varphi_1)$$

Therefore

$$\rank(\varphi_1) = 1 \neq 2 = \rank(\varphi_0)$$

mucciolo
  • 3,008
  • So $\varphi_0$ is a map from $V_0$ to $V_0$ and $\varphi_1$ is a map from $V_1$ to $V_1$. If $V_0\neq V_1$, do you actually mean $\ker(\varphi_0 \circ\theta)\subset \ker(\varphi_1^{n-1})$? – qinr Apr 25 '20 at 23:36
  • Yes but I am indiscriminately looking at $\varphi_0$ as the map $V_0 \oplus {0}^k \to V_0 \oplus {0}^k : (x, 0) \mapsto \varphi(x, 0)$ since $V_0 \oplus {0}^k \cong V_0$ and $\varphi$ preserves $V_0$; and analogously for $\varphi_1$ . Also I don't see why we should be concerned with the case when $V_0 \neq V_1$ since both have the same finite dimension, so $V_0 \cong \mathbb{R}^k \cong V_1$, and then they can be reduced to the case of $\mathbb{R}^k \oplus \mathbb{R}^k$ with the canonical basis, thus $\theta$ preserving their kernel. – mucciolo Apr 26 '20 at 00:39